ICSE Class 9 Maths Selina Solutions Chapter 4:
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ICSE Class 9 Maths Selina Solutions Chapter 4 Expansion Overview
In mathematics, the expansion of a product of sums is expressed as the sum of its products using the fact that multiplication divides addition - the reverse process of trying to write an expanded polynomial as a product is called polynomial factors. "Expand" means to remove ( ) ... but we have to do it right ( ) are called "parentheses" or "brackets". Anything inside ( ) should be treated as "package". So to say: tell everything inside the "package".
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ICSE Class 9 Maths Selina Solutions Chapter 4 PDF
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ICSE Class 9 Maths Selina Solutions Chapter 4 PDF
ICSE Class 9 Maths Selina Solutions Chapter 4 Expansion
Here we have provided ICSE Class 9 Maths Selina Solutions Chapter 4 Expansion for the ease of students so that they can prepare better for their ICSE Class 9 Maths Exam.
ICSE Class 9 Maths Selina Solutions Chapter 4 Exercise 4(A)
1. Find the square of:
(i) 2a + b
(ii) 3a + 7b
(iii) 3a – 4b
(iv) 3a/2b – 2b/3a
Solution:
Using the identities,
(a + b)
2
= a
2
+ b
2
+ 2ab and
(a – b)
2
= a
2
+ b
2
– 2ab
(i) (2a + b)
2
= (2a)
2
+ b
2
+ 2(2a)(b)
= 4a
2
+ b
2
+ 4ab
(ii) (3a + 7b)
2
= (3a)
2
+ (7b)
2
+ 2(3a)(7b)
= 9a
2
+ 49b
2
+ 42ab
(iii) (3a – 4b)
2
= (3a)
2
+ (4b)
2
– 2(3a)(4b)
= 9a
2
+ 16b
2
– 24ab
(iv) (3a/2b – 2b/3a)
2
= (3a/2b)
2
+ (2b/3a)
2
– 2(3a/2b)(2b/3a)
= 9a
2
/4b
2
+ 4b
2
/9a
2
– 2
2. Use identities to evaluate:
(i) (101)
2
(ii) (502)
2
(iii) (97)
2
(iv) (998)
2
Solution:
Using the identities,
(a + b)
2
= a
2
+ b
2
+ 2ab and
(a – b)
2
= a
2
+ b
2
– 2ab
(i) (101)
2
= (100 + 1)
2
= 100
2
+ 1
2
+ 2×100×1
= 10000 + 1 + 200
= 10201
(ii) (502)
2
= (500 + 2)
2
= 500
2
+ 2
2
+ 2×500×2
= 250000 + 4 + 2000
= 252004
(iii) (97)
2
= (100 – 3)
2
= 100
2
+ 3
2
– 2×100×3
= 10000 + 9 – 600
= 9409
(iv) (998)
2
= (1000 – 2)
2
= 1000
2
+ 2
2
– 2×1000×2
= 100000 + 4 – 4000
= 996004
3. Evalute:
(i) (7x/8 + 4y/5)
2
(ii) (2x/7 – 7y/4)
2
Solution:
Using the identities,
(a + b)
2
= a
2
+ b
2
+ 2ab and
(a – b)
2
= a
2
+ b
2
– 2ab
(i) We have, (7x/8 + 4y/5)
2
= (7x/8)
2
+ (4y/5)
2
+ 2 × (7x/8) × (4y/5)
= 49x
2
/64 + 16y
2
/25 + 7xy/5
(ii) We have, (2x/7 – 7y/4)
2
= (2x/7)
2
+ (7y/4)
2
– 2 × (2x/7) × (7y/4)
= 4x
2
/49 + 49y
2
/16 – xy
4. Evaluate:
(i) (a/2b + 2b/a)
2
– (a/2b – 2b/a)
2
– 4
(ii) (4a + 3b)
2
– (4a – 3b)
2
+ 48ab
Solution:
Using the identities,
(a + b)
2
= a
2
+ b
2
+ 2ab and
(a – b)
2
= a
2
+ b
2
– 2ab
(i) Given expression,
On expanding the term using the identity, we have
(a/2b + 2b/a)
2
= (a/2b)
2
+ (2b/a)
2
+ 2 × (a/2b) × (2b/a)
= a
2
/4b + 4b
2
/a
2
+ 2
Next, expanding the second term using the identity, we have
(a/2b – 2b/a)
2
= (a/2b)
2
+ (2b/a)
2
– 2 × (a/2b) × (2b/a)
= a
2
/4b + 4b
2
/a
2
– 2
Now, using these results in the given expression
(a/2b + 2b/a)
2
– (a/2b – 2b/a)
2
– 4 = (a
2
/4b + 4b
2
/a
2
+ 2) – (a
2
/4b + 4b
2
/a
2
– 2) – 4
= a
2
/4b + 4b
2
/a
2
+ 2 – a
2
/4b – 4b
2
/a
2
+ 2 – 4
= 0
(ii) Given expression, (4a + 3b)
2
– (4a – 3b)
2
+ 48ab
On expanding the term using the identity, we have
(4a + 3b)
2
= (4a)
2
+ (3b)
2
+ 2 × (4a) × (3b)
= 16a
2
+ 9b
2
+ 24ab
Next, expanding the second term using the identity, we have
(4a – 3b)
2
= (4a)
2
+ (3b)
2
– 2 × (4a) × (3b)
= 16a
2
+ 9b
2
– 24ab
Now, using these results in the given expression
(4a + 3b)
2
– (4a – 3b)
2
+ 48ab = (16a
2
+ 9b
2
+ 24ab) – (16a
2
+ 9b
2
– 24ab) + 48ab
= 16a
2
+ 9b
2
+ 24ab – 16a
2
– 9b
2
+ 24ab + 48ab
= 96ab
5. If a + b = 7 and ab = 10; find a – b.
Solution:
Using the identities,
(a + b)
2
= a
2
+ b
2
+ 2ab and
(a – b)
2
= a
2
+ b
2
– 2ab
Rewriting the above equation as
(a – b)
2
= a
2
+ b
2
+ 2ab – 4ab
= (a + b)
2
– 4ab … (i)
We have,
a + b = 7 and ab = 10
So, using these in equation (i), we get
(a – b)
2
= (7)
2
– (4 × 10)
= 49 – 40
= 9
Then,
(a – b) = √9
= ±3
6. If a – b = 7 and ab = 18; find a + b.
Solution:
Using the identities,
(a – b)
2
= a
2
+ b
2
– 2ab and
(a + b)
2
= a
2
+ b
2
+ 2ab
Rewriting the above equation as
(a + b)
2
= a
2
+ b
2
– 2ab + 4ab
= (a – b)
2
+ 4ab … (i)
We have,
a – b = 7 and ab = 18
So, using these in equation (i), we get
(a + b)
2
= (7)
2
+ (4 × 18)
= 49 + 72
= 121
Then,
(a + b) = √121
= ±11
7. If x + y = 7/2 and xy = 5/2; find:
(i) x – y
(ii) x
2
– y
2
Solution:
Using the identities,
(x + y)
2
= x
2
+ y
2
+ 2xy and
(x – y)
2
= x
2
+ y
2
– 2xy
Rewriting the above equation as
(x – b)
2
= x
2
+ y
2
+ 2xy – 4xy
= (x + y)
2
– 4xy … (1)
We have,
x + y = 7/2 and xy = 5/2
So, using these in equation (1), we get
(x – y)
2
= (7/2)
2
– (4 × 5/2)
= 49/4 – 10
= 9/4
Then,
(x – y) = √(9/4)
= ± 3/2 … (2)
(ii) We know that,
x
2
– y
2
= (x + y) (x – y)
Substituting values in RHS using given and (2), we get
x
2
– y
2
= (7/2) (± 3/2)
= ± 21/4
8. If a – b = 0.9 and ab = 0.36; find:
(i) a + b
(ii) a
2
– b
2
Solution:
Using the identities,
(a – b)
2
= a
2
+ b
2
– 2ab and
(a + b)
2
= a
2
+ b
2
+ 2ab
Rewriting the above equation as
(a + b)
2
= a
2
+ b
2
– 2ab + 4ab
= (a – b)
2
+ 4ab … (1)
We have,
a – b = 0.9 and ab = 0.36
So, using these in equation (1), we get
(a + b)
2
= (0.9)
2
+ (4 × 0.36)
= 0.81 + 1.44
= 2.25
Then,
(a + b) = √2.25
= ±1.5 … (2)
(ii) We know that,
a
2
– b
2
= (a + b) (a – b)
Substituting values in RHS using given and (2), we get
a
2
– b
2
= (±1.5) (0.9)
= ± 1.35
9. If a – b = 4 and a + b = 6; find
(i) a
2
+ b
2
(ii) ab
Solution:
Given, a – b = 4 and a + b = 6
We know that,
(a – b)
2
= a
2
+ b
2
– 2ab and
(a + b)
2
= a
2
+ b
2
+ 2ab
Now, rewriting the above equation as
(a + b)
2
= a
2
+ b
2
– 2ab + 4ab
⇒ (a + b)
2
= (a – b)
2
+ 4ab
Substituting the values in the above equation, we get
(6)
2
= (4)
2
+ 4ab
36 = 16 + 4ab
4ab = 36 – 16
4ab = 20
ab = 20/4
(ii) Thus, ab = 5
Now, in the identity: (a + b)
2
= a
2
+ b
2
+ 2ab
(a + b)
2
= (a
2
+ b
2
) + 2ab
Let’s substitute the values of the known terms,
(6)
2
= (a
2
+ b
2
) + 2 × (5)
36 = (a
2
+ b
2
) + 10
a
2
+ b
2
= 36 – 10
(i) Thus, a
2
+ b
2
= 26
10. If a + 1/a = 6 and a ≠ 0 find:
(i) a – 1/a
(ii) a
2
– 1/a
2
Solution:
Using the identities,
(a – b)
2
= a
2
+ b
2
– 2ab and
(a + b)
2
= a
2
+ b
2
+ 2ab
(i) Now,
(a + 1/a)
2
= a
2
+ 1/a
2
+ (2 × a × 1/a)
= a
2
+ 1/a
2
+ 2 … (1)
Substituting the value of (a + 1/a) in the equation (1), we get
6
2
= a
2
+ 1/a
2
+ 2
36 = a
2
+ 1/a
2
+ 2
a
2
+ 1/a
2
= 36 – 2 = 34 … (2)
Similarly,
(a – 1/a)
2
= a
2
+ 1/a
2
– (2 × a × 1/a)
= (a
2
+ 1/a
2
) – 2
= 34 – 2 … [From (2)]
= 32
⇒ (a – 1/a)
2
= 32
a – 1/a = ±√32
= ±4√2 … (3)
Thus, a – 1/a = ±4√2
(ii) We know that,
a
2
– 1/a
2
= (a – 1/a) (a + 1/a)
Using the given and (3) in the above equation,
a
2
– 1/a
2
= (±4√2) (6)
= ±24√2
Thus, a
2
– 1/a
2
= ±24√2
11. If a – 1/a = 8 and a ≠ 0, find:
(i) a + 1/a
(ii) a
2
– 1/a
2
Solution:
Using the identities,
(a – b)
2
= a
2
+ b
2
– 2ab and
(a + b)
2
= a
2
+ b
2
+ 2ab
(i) Now,
(a – 1/a)
2
= a
2
+ 1/a
2
– (2 × a × 1/a)
= a
2
+ 1/a
2
– 2 … (1)
Substituting the value of (a + 1/a) in the equation (1), we get
8
2
= a
2
+ 1/a
2
– 2
64 = a
2
+ 1/a
2
– 2
a
2
+ 1/a
2
= 64 + 2 = 66 … (2)
Similarly,
(a + 1/a)
2
= a
2
+ 1/a
2
+ (2 × a × 1/a)
= (a
2
+ 1/a
2
) + 2
= 66 + 2 … [From (2)]
= 68
⇒ (a + 1/a)
2
= 68
a + 1/a = √68
= ±2√17 … (3)
Thus, a + 1/a = ±2√17
(ii) We know that,
a
2
– 1/a
2
= (a – 1/a) (a + 1/a)
Using the given and (3) in the above equation,
a
2
– 1/a
2
= (8) (±2√17)
= ±16√17
Thus, a
2
– 1/a
2
= ±16√17
12. If a
2
– 3a + 1 = 0, and a ≠ 0; find:
(i) a + 1/a
(ii) a
2
+ 1/a
2
Solution:
(i) Given equation,
a
2
– 3a + 1 = 0
a
2
+ 1 = 3a
(a
2
+ 1)/a = 3
⇒ a + 1/a = 3 … (1)
(ii) We know that,
(a + b)
2
= a
2
+ b
2
+ 2ab
Now,
(a + 1/a)
2
= a
2
+ 1/a
2
+ 2(a)(1/a)
= a
2
+ 1/a
2
+ 2
Using (1) in the above equation, we get
(3)
2
= a
2
+ 1/a
2
+ 2
9 = a
2
+ 1/a
2
+ 2
a
2
+ 1/a
2
= 9 – 2
Thus, a
2
+ 1/a
2
= 7
13. If a
2
– 5a – 1 = 0 and a ≠ 0; find:
(i) a – 1/a
(ii) a + 1/a
(iii) a
2
– 1/a
2
Solution:
(i) Given, a
2
– 5a – 1 = 0
Rewriting the equation, we get
a
2
– 1 = 5a
(a
2
– 1)/a = 5
Hence, a – 1/a = 5 … (1)
(ii) We know that,
(a + 1/a)
2
= a
2
+ 1/a
2
+ 2
Manipulating the above as,
(a + 1/a)
2
= a
2
+ 1/a
2
– 2 + 4
(a + 1/a)
2
= (a – 1/a)
2
+ 4
Now, using (1) in the above
(a + 1/a)
2
= (5)
2
+ 4
(a + 1/a)
2
= 25 + 4 = 29
Hence, a + 1/a = ±√29 … (2)
(iii) We know that,
a
2
– 1/a
2
= (a + 1/a) (a – 1/a)
Now, using (1) and (2) in the above equation, we get
a
2
– 1/a
2
= (5) x (±√29)
Hence, a
2
– 1/a
2
= ±5√29
14. If 3a + 4b = 16 and ab = 4; find the value of 9a
2
+ 16b
2
.
Solution:
Given, 3a + 4b = 16 and ab = 4
Required to find: value of 9a
2
+ 16b
2
We know that,
(a + b)
2
= a
2
+ b
2
+ 2ab
Now, the square of (3a + 4b) will be
(3a + 4b)
2
= (3a)
2
+ (4b)
2
+ 2 × (3a) × (4b)
= 9a
2
+ 4b
2
+ 24ab
And, given 3a + 4b = 16
So, by squaring on both the sides
(3a + 4b)
2
= 16
2
9a
2
+ 4b
2
+ 24ab = 256
9a
2
+ 4b
2
+ 24(4) = 256 [Given ab = 4]
9a
2
+ 4b
2
= 256 – 96
⇒ 9a
2
+ 4b
2
= 160
15. The number a is 2 more than the number b. If the sum of the squares of a and b is 34, then find the product of a and b.
Solution:
Given, a is 2 more than b
⇒ a = b + 2
And, sum of squares of a and b is 34
⇒ a
2
+ b
2
= 34
Let’s replace a = (b + 2) in the above equation and solve for b
Then,
(b + 2)
2
+ b
2
= 34
2b
2
+ 4b – 30 = 0
b
2
+ 2b – 15 = 0
(b + 5) (b – 3) = 0
So,
b = -5 or 3
Now,
For b = -5, a =-5 + 2 = -3
For b = 3, a = 3 + 2 = 5
Thus, the product of a and b is 15 in both cases.
16. The difference between two positive numbers is 5 and the sum of their squares is 73. Find the product of these numbers.
Solution:
Let’s assume the two positive numbers as a and b
Given, the difference between them is 5 and the sum of their squares is 73
So, we have
a – b = 5 … (i) and
a
2
+ b
2
= 73 … (ii)
On squaring (i) on both sides, we get
(a – b)
2
= 5
2
(a
2
+ b
2
)
– 2ab = 25
73
– 2ab = 25 … [Using (ii), given]
So,
2ab = 73 – 25 = 48
ab = 24
Therefore, the product of numbers is 24.
Exercise 4(B)
1. Find the cube of:
(i) 3a – 2b
(ii) 5a + 3b
(iii) 2a + 1/2a
(iv) 3a – 1/a (a
≠ 0)
Solution:
Using the identities,
(a + b)
3
= a
3
+ 3ab (a + b) + b
3
and
(a – b)
3
= a
3
– 3ab (a – b) + b
3
(i) (3a – 2b)
3
= (3a)
3
– 3 × 3a × 2b (3a – 2b) – (2b)
3
= 27a
3
– 18ab (3a – 2b) – 8b
3
= 27a
3
– 54a
2
b + 36ab
2
– 8b
3
(ii) (5a + 3b)
3
= (5a)
3
+ 3 × 5a × 3b (5a + 3b) + (3b)
3
= 125a
3
+ 45ab (5a + 3b) + 27b
3
= 125a
3
+ 225a
2
b + 135ab
2
+ 27b
3
(iii) (2a + 1/2a)
3
= (2a)
3
+ 3 × 2a × 1/2a (2a + 1/2a) + (1/2a)
3
= 8a
3
+ 3 (2a + 1/2a) + 1/8a
3
= 8a
3
+ 6a + 3/2a + 1/8a
3
(iv) (3a – 1/a)
3
= (3a)
3
– 3 × 3a × 1/a (3a – 1/a) – (1/a)
3
= 27a
3
– 9 (3a – 1/a) – 1/a
3
= 27a
3
– 27a + 9a – 1/a
3
2. If a
2
+ 1/a
2
= 47 and a ≠ 0 find:
(i) a + 1/a
(ii) a
3
+ 1/a
3
Solution:
(i) Given, a
2
+ 1/a
2
= 47
We know that,
(a + 1/a)
2
= a
2
+ 1/a
2
+ 2 x a x 1/a
= (a
2
+ 1/a
2
) + 2
= 47 + 2
= 49
So,
a + 1/a = √49
= ±7 … (1)
(ii) Using the identity
(a + b)
3
= a
3
+ 3ab (a + b) + b
3
Now,
(a + 1/a)
3
= a
3
+ 1/a
3
+ 3(a + 1/a)
a
3
+ 1/a
3
= (a + 1/a)
3
– 3(a + 1/a)
= (±7)
3
– 3(±7) … [From (1)]
= ±343 – ±21
Hence, a
3
+ 1/a
3
= ±322
3. If a
2
+ 1/a
2
= 18; a ≠ 0 find:
(i) a – 1/a
(ii) a
3
– 1/a
3
Solution:
(i) Given, a
2
+ 1/a
2
= 18
Using the identity (a + b)
2
= a
2
+ b
2
+ 2ab
Now,
(a – 1/a)
2
= a
2
+ 1/a
2
– 2(a)(1/a)
= (a
2
+ 1/a
2
) – 2
= 18 – 2
= 16
Hence,
a – 1/a = √16
= ±4 … (1)
(ii) Using the identity,
(a – b)
3
= a
3
– 3ab (a – b) + b
3
Now,
(a – 1/a)
3
= a
3
– 3a(1/a) (a – 1/a) + (1/a)
3
= a
3
– 3 (a – 1/a) + 1/a
3
a
3
+ 1/a
3
= (a – 1/a)
3
+ 3 (a – 1/a)
= (±4)
3
+ 3(±4)
= ±64 ± 12
Hence,
a
3
+ 1/a
3
= ±76
4. If a + 1/a = p and a ≠ 0; then show that:
a
3
+ 1/a
3
= p (p
2
– 3)
Solution:
Given, a + 1/a = p … (1)
Now, cubing on both sides
(a + 1/a)
3
= p
3
a
3
+ 1/a
3
+ 3(a + 1/a) = p
3
a
3
+ 1/a
3
= p
3
– 3(a + 1/a)
= p
3
– 3(p) [From (1)]
= p (p
2
– 3)
– Hence proved
5. If a + 2b = 5; then show that:
a
3
+ 8b
3
+ 30ab = 125.
Solution:
Given, a + 2b = 5
Let’s cube it on both sides,
(a + 2b)
3
= 5
3
a
3
+ 3(a)(2b)(a + 2b) + (2b)
3
= 125
a
3
+ 6ab(a + 2b) + 8b
3
= 125
a
3
+ 8b
3
= 125 – 6ab (a + 2b)
= 125 – 6ab (5) … [Given]
= 125 – 30ab
So,
a
3
+ 8b
3
+ 30ab = 125
– Hence showed
6. If (a + 1/a)
2
= 3 and a ≠ 0, then show: a
3
+ 1/a
3
= 0.
Solution:
Given, (a + 1/a)
2
= 3
⇒ a + 1/a = ±√3 … (1)
We know the identity,
(a + 1/a)
3
= a
3
+ 1/a
3
+ 3(a + 1/a)
a
3
+ 1/a
3
= (a + 1/a)
3
– 3(a + 1/a)
= (±√3)
3
– 3(±√3)
= ±3√3 – (±3√3)
= 0
Thus, a
3
+ 1/a
3
= 0
7. If a + 2b + c = 0; then show that:
a
3
+ 8b
3
+ c
3
= 6abc
Solution:
We have, a + 2b + c = 0
a + 2b = -c
Now, on cubing it on both sides we get
(a + 2b)
3
= (-c)
3
a
3
+ (2b)
3
+ 3(a)(2b)(a + 2b) = -c
3
a
3
+ 8b
3
+ 6ab (a + 2b) = -c
3
a
3
+ 8b
3
+ 6ab (-c) = -c
3
a
3
+ 8b
3
– 6abc = -c
3
Hence,
a
3
+ 8b
3
+ c
3
= 6abc
8. Use property to evaluate:
(i) 13
3
+ (-8)
3
+ (-5)
3
(ii) 7
3
+ 3
3
+ (-10)
3
(iii) 9
3
– 5
3
– 4
3
(iv) 38
3
+ (-26)
3
+ (-12)
3
Solution:
The property is if a + b + c = 0 then
a
3
+ b
3
+ c
3
= 3abc
Now,
(i) a = 13, b = -8 and c = -5
⇒ 13
3
+ (-8)
3
+ (-5)
3
= 3(13) (-8) (-5) … [Since, 13 + (-8) + (-5) = 0]
= 1560
(ii) a = 7, b = 3, c = -10
⇒ 7
3
+ 3
3
+ (-10)
3
= 3(7) (3) (-10) … [Since, 7 + 3 + (-10) = 0]
= -630
(iii)a = 9, b = -5, c = -4
⇒ 9
3
– 5
3
– 4
3
= 9
3
+ (-5)
3
+ (-4)
3
… [Since, 9 + (-5) + (-4) = 0]
= 3(9) (-5) (-4) = 540
(iv) a = 38, b = -26, c = -12
⇒ 38
3
+ (-26)
3
+ (-12)
3
= 3(38) (-26) (-12) … [Since, 38 + (-26) + (-12) = 0]
= 35568
9. If a ≠ 0 and a – 1/a = 3; find:
(i) a
2
+ 1/a
2
(ii) a
3
– 1/a
3
Solution:
(i) We have, a – 1/a = 3
On squaring on both sides, we get
(a – 1/a)
2
= 3
2
a
2
+ 1/a
2
– 2 = 9
a
2
+ 1/a
2
= 9 + 2
Hence,
a
2
+ 1/a
2
= 11
(ii) We have, a – 1/a = 3
On cubing on both sides, we get
(a – 1/a)
3
= 3
3
a
2
– 1/a
3
– 3(a – 1/a) = 27
a
2
– 1/a
3
= 27 + 3(a – 1/a)
= 27 + 3(3)
= 27 + 9
Hence,
a
3
– 1/a
3
= 36
10. If a ≠ 0 and a – 1/a = 4; find:
(i) a
2
+ 1/a
2
(ii) a
4
+ 1/a
4
(iii) a
3
– 1/a
3
Solution:
(i) We have, a – 1/a = 4 … (a)
On squaring it on both sides, we get
(a – 1/a)
2
= 4
2
a
2
+ 1/a
2
– 2(a)(1/a) = 16
a
2
+ 1/a
2
– 2 = 16
a
2
+ 1/a
2
= 16 + 2 = 18 … (1)
Hence, a
2
+ 1/a
2
= 18
(ii) Now, we know that
a
4
+ 1/a
4
= (a
2
+ 1/a
2
)
2
– 2
= 18
2
– 2 … [From (1)]
= 324 – 2
Hence, a
4
+ 1/a
4
= 322
(iii) On cubing (i) on both sides, we get
(a – 1/a)
3
= 4
3
a
3
– 1/a
3
– 3(a – 1/a) = 64
a
3
– 1/a
3
= 64 + 3(a – 1/a)
= 64 + 3(4) … [Given]
= 64 + 12
Hence, a
3
– 1/a
3
= 76
11. If x ≠ 0 and x + 1/x = 2; then show that:
x
2
+ 1/x
2
= x
3
+ 1/x
3
= x
4
+ 1/x
4
Solution:
We have, x + 1/x = 2
We know that,
(x + 1/x)
2
= x
2
+ 1/x
2
+ 2
(2)
2
= x
2
+ 1/x
2
+ 2
x
2
+ 1/x
2
= 4 – 2
= 2 … (i)
Next, calculating
(x + 1/x)
3
= x
3
+ 1/x
3
+ 3(x + 1/x)
(2)
3
= x
3
+ 1/x
3
+ 3(2)
x
3
+ 1/x
3
= 2
3
– 3(2)
= 8 – 6
= 2 … (ii)
Next, we know that
x
4
+ 1/x
4
= (x
2
+ 1/x
2
) – 2
= 2
2
– 2 … [From (i)]
= 4 – 2
= 2 … (iii)
Therefore, from (i), (ii) and (iii) we have
x
2
+ 1/x
2
= x
3
+ 1/x
3
= x
4
+ 1/x
4
12. If 2x – 3y = 10 and xy = 16; find the value of 8x
3
– 27y
3
.
Solution:
Given,
2x – 3y = 10 … (i) and
xy = 16 … (ii)
Now, on cubing (i) on both sides
(2x – 3y)
3
= 10
3
(2x)
3
– 3(2x)(3y) (2x – 3y) – (3y)
3
= 1000 []
8x
3
– 18(xy) (2x – 3y) – 27y
3
= 1000
8x
3
– 18 × 16 × 10 – 27y
3
= 1000
8x
3
– 2880 – 27y
3
= 1000
8x
3
– 27y
3
= 1000 + 2880
8x
3
– 27y
3
= 3880
13. Expand:
(i) (3x + 5y + 2z) (3x – 5y + 2z)
(ii) (3x – 5y – 2z) (3x – 5y + 2z)
Solution:
(i) We have, (3x + 5y + 2z) (3x – 5y + 2z)
= {(3x + 2z) + (5y)} {(3x + 2z) – (5y)} … [By grouping]
= (3x + 2z)
2
– (5y)
2
… [As (a + b) (a – b) = a
2
– b
2
]
= 9x
2
+ 4z
2
+ (2 × 3x × 2z) – 25y
2
= 9x
2
+ 4z
2
+ 12xz – 25y
2
= 9x
2
+ 4z
2
– 25y
2
+ 12xz
(ii) We have, (3x – 5y – 2z) (3x – 5y + 2z)
= {(3x – 5y) – (2z)} {(3x – 5y) + (2z)} … [By grouping]
= (3x – 5y)
2
– (2z)
2
… [As (a + b) (a – b) = a
2
– b
2
]
= 9x
2
+ 25y
2
– 2 × 3x × 5y – 4z
2
= 9x
2
+ 25y
2
– 30xy – 4z
2
= 9x
2
+25y
2
– 4z
2
– 30xy
14. The sum of two numbers is 9 and their product is 20. Find the sum of their
(i) Squares (ii) Cubes
Solution:
Given, the sum of two numbers is 9 and their product is 20
Let’s assume the numbers to ‘a’ and ‘b’
So, we have
a + b = 9 … (1) and
ab = 20 … (2)
Now,
On squaring (1) on both sides gives, we get
(a + b)
2
= 9
2
a
2
+ b
2
+ 2ab = 81
a
2
+ b
2
+ 2(20) = 81 … [From (2)]
a
2
+ b
2
+ 40 = 81
a
2
+ b
2
= 81 – 40 = 41
(i) Hence, the sum of their squares is 41
Next,
On cubing (1) on both sides, we get
(a + b)
3
= 9
3
a
3
+ b
3
+ 3ab (a + b) = 729
a
3
+ b
3
+ 3 × (20) × (9) = 729 … [From (1) and (2)]
a
3
+ b
3
= 729 – 540 = 189
(ii) Hence, the sum of their cubes is 189.
15. Two positive numbers x and y are such that x > y. If the difference of these numbers is 5 and their product is 24, find:
(i) Sum of these numbers
(ii) Difference of their cubes
(iii) Sum of their cubes.
Solution:
Given x – y = 5 and xy = 24 (x>y)
(x + y)
2
= (x – y)
2
+ 4xy = 25 + 96 = 121
So, x + y = 11; sum of these numbers is 11.
Cubing on both sides gives
(x – y)
3
= 5
3
x
3
– y
3
– 3xy(x – y) = 125
x
3
– y
3
– 72(5) = 125
x
3
– y
3
= 125 + 360 = 485
So, difference of their cubes is 485.
Cubing both sides, we get
(x + y)
3
= 11
3
x
3
+ y
3
+ 3xy(x + y) = 1331
x
3
+ y
3
= 1331 – 72(11) = 1331 – 792 = 539
So, sum of their cubes is 539.
16. If 4x
2
+ y
2
= a and xy = b, find the value of 2x + y.
Solution:
Given, xy = b … (i) and 4x
2
+ y
2
= a … (ii)
Now,
(2x + y)
2
= (2x)
2
+ 4xy + y
2
= (4x
2
+ y
2
) + 4xy
= a + 4b … [Using (i) and (ii)]
Hence,
2x + y = ±√(a + 4b)
Exercise 4(C)
1. Expand:
(i) (x + 8) (x + 10)
(ii) (x + 8) (x – 10)
(iii) (x – 8) (x + 10)
(iv) (x – 8) (x – 10)
Solution:
Using the identity, (x + a) (x + b) = x
2
+ (a + b) x + ab
(i) We have, (x + 8) (x + 10)
= x
2
+ (8 + 10) x + 8 × 10
= x
2
+ 18x + 80
(ii) We have, (x + 8) (x – 10)
= x
2
+ (8 – 10) x + 8 × (-10)
= x
2
– 2x – 80
(iii) (ii) We have, (x – 8) (x + 10)
= x
2
+ (-8 + 10) x + (-8) × 10
= x
2
+ 2x – 80
(iv) We have, (x – 8) (x – 10)
= x
2
+ (-8 – 10) x + (-8) × (-10)
= x
2
– 18x + 80
2. Expand:
(i) (2x – 1/x) (3x + 2/x)
(ii) (3a + 2/b) (2a – 3/b)
Solution:
(i) We have, (2x – 1/x) (3x + 2/x)
= (2x)(3x) + (2x)(2/x) – (1/x)(3x) – (1/x)(2/x)
= 6x
2
+ 4 – 3 – 2/x
2
= 6x
2
+ 1 – 2/x
2
(ii) We have, (3a + 2/b) (2a – 3/b)
= (3a)(2a) – (3a)(3/b) + (2/b)(2a) – (2/b)(3/b)
= 6a
2
– 9a/b + 4a/b – 6/b
2
= 6a
2
– 5a/b – 6/b
2
3. Expand:
(i) (x + y – z)
2
(ii) (x – 2y + 2)
2
(iii) (5a – 3b + c)
2
(iv) (5x – 3y – 2)
2
(v) (x – 1/x + 5)
2
Solution:
(i) (x + y – z)
2
= x
2
+ y
2
+ z
2
+ 2(x)(y) – 2(y)(z) – 2(z)(x)
= x
2
+ y
2
+ z
2
+ 2xy – 2yz – 2zx
(ii) (x – 2y + 2)
2
= x
2
+ (-2y)
2
+ 2
2
+ 2(x)(-2y) + 2(-2y)(2) + 2(2)(x)
= x
2
+ 4y
2
+ 4 – 4xy – 8y + 4x
(iii) (5a – 3b + c)
2
= (5a)
2
+ (-3b)
2
+ c
2
+ 2(5a)(-3b) + 2(-3b)(c) + 2(c)(5a)
= 25a
2
+ 9b
2
+ c
2
– 30ab – 6bc + 10ac
(iv) (5x – 3y – 2)
2
= (5x)
2
+ (-3y)
2
+ (-2)
2
+ 2(5x)(-3y) + 2(-3y)(-2) + 2(-2)(5x)
= 25x
2
+ 9y
2
+ 4 – 30xy + 12y – 20x
(v) (x – 1/x + 5)
2
= (x)
2
+ (-1/x)
2
+ (5)
2
+ 2(x)(-1/x) + 2(-1/x)(5) + 2(5)(x)
= x
2
+ 1/x
2
+ 25 – 2 – 10/x + 10x
= x
2
+ 1/x
2
+ 23 – 10/x + 10x
4. If a + b + c = 12 and a
2
+ b
2
+ c
2
= 50; find ab + bc + ca.
Solution:
Given, a + b + c = 12 and a
2
+ b
2
+ c
2
= 50
We know that,
(a + b + c)
2
= a
2
+ b
2
+ c
2
+ 2(ab + bc + ca)
12
2
= 50 + 2(ab + bc + ca)
144 = 50 + 2(ab + bc + ca)
ab + bc + ca = (144 – 50)/ 2
= 94/2
Thus,
ab + bc + ca = 47
5. If a
2
+ b
2
+ c
2
= 35 and ab + bc + ca = 23; find a + b + c.
Solution:
Given, a
2
+ b
2
+ c
2
= 35 and ab + bc + ca = 23
We know that,
(a + b + c)
2
= (a
2
+ b
2
+ c
2
) + 2(ab + bc + ca)
(a + b + c)
2
= 35 + 2(23)
(a + b + c)
2
= 35 + 46
(a + b + c)
2
= 81
(a + b + c) = ±√81
Thus,
a + b + c = ±9
6. If a + b + c = p and ab + bc + ca = q; find a
2
+ b
2
+ c
2
.
Solution:
Given, a + b + c = p and ab + bc + ca = q
We know that,
(a + b + c)
2
= (a
2
+ b
2
+ c
2
) + 2(ab + bc + ca)
(p)
2
= (a
2
+ b
2
+ c
2
) + 2(q)
⇒ a
2
+ b
2
+ c
2
= p
2
– 2q
7. If a
2
+ b
2
+ c
2
= 50 and ab + bc + ca = 47, find a + b + c.
Solution:
Given, a
2
+ b
2
+ c
2
= 50 and ab + bc + ca = 47
We know that,
(a + b + c)
2
= (a
2
+ b
2
+ c
2
) + 2(ab + bc + ca)
(a + b + c)
2
= 50 + 2(47)
(a + b + c)
2
= 50 + 94
= 144
⇒ (a + b + c) = √144
Thus,
a + b + c = ±12
8. If x + y – z = 4 and x
2
+ y
2
+ z
2
= 30, then find the value of xy – yz – zx.
Solution:
Given, x + y – z = 4 and x
2
+ y
2
+ z
2
= 30
We know that,
(x + y – z)
2
= x
2
+ y
2
+ z
2
+ 2(xy – yz – zx)
4
2
= 30 + 2(xy – yz – zx)
16 – 30 = 2(ab + bc + ca)
xy – yz – zx = -14/ 2
Thus,
xy – yz – zx = -7
Exercise 4(D)
1. If x + 2y + 3z = 0 and x
3
+ 4y
3
+ 9z
3
= 18xyz; evaluate:
Solution:
Given, x
3
+ 4y
3
+ 9z
3
= 18xyz and x + 2y + 3z = 0
So,
x + 2y = – 3z, 2y + 3z = -x and 3z + x = -2y
Now,
2. If a + 1/a = m and a ≠ 0; find in terms of ‘m’; the value of:
(i) a – 1/a
(ii) a
2
– 1/a
2
Solution:
(i) Given, a + 1/a = m
On squaring on both sides, we get
(a + 1/a)
2
= m
2
a
2
+ 1/a
2
+ 2 = m
2
a
2
+ 1/a
2
= m
2
– 2 … (1)
Now, consider the expansion
(a – 1/a)
2
= a
2
+ 1/a
2
– 2
= m
2
– 2 – 2 … [From (1)]
= m
2
– 4
So,
(a – 1/a) = ±√(m
2
– 4) … (2)
(ii) We know that,
a
2
– 1/a
2
= (a – 1/a) (a + 1/a)
= m [±√(m
2
– 4)]
= ±m√(m
2
– 4)
3. In the expansion of (2x
2
– 8) (x – 4)
2
; find the value of
(i) coefficient of x
3
(ii) coefficient of x
2
(iii) constant term
Solution:
We have, (2x
2
– 8) (x – 4)
2
= (2x
2
– 8) (x
2
– 2 × 4 × x + 4
2
)
= (2x
2
– 8) (x
2
– 8x + 16)
= 2x
2
(x
2
– 8x + 16) – 8(x
2
– 8x + 16)
= 4x
4
– 16x
3
+ 32x
2
– 8x
2
+ 64x – 128
= 4x
4
– 16x
3
+ 24x
2
+ 64x – 128
Now,
(i) coefficient of x
3
= -16
(ii) coefficient of x
2
= 24
(iii) constant term = -128
4. If x > 0 and x
2
+ 1/9x
2
= 25/36. Find: x
3
+ 1/27x
3
Solution:
Given, x
2
+ 1/9x
2
= 25/36 … (1)
Now, consider the expansion
(x + 1/3x)
2
= x
2
+ (1/3x)
2
+ (2 × x × 1/3x)
= (x
2
+ 1/9x
2
) + 2/3
= 25/36 + 2/3 … [From (1)]
= 49/36
So,
(x + 1/3x) = ±√(49/36)
= ±7/6 … (2)
Now, consider the expansion
(x + 1/3x)
3
= x
3
+ (1/3x)
3
+ 3(x + 1/3x)
(7/6)
3
= x
3
+ (1/3x)
3
+ 3(7/6) …… [From (2)]
343/216 = x
3
+ 1/27x
3
+ 21/6
x
3
+ 1/27x
3
= 343/216 – 21/6
= (343 – 252)/216
= 91/216
Thus, x
3
+ 1/27x
3
= 91/216
5. If 2(x
2
+ 1) = 5x, find:
(i) x – 1/x
(ii) x
3
– 1/x
3
Solution:
(i) Given, 2(x
2
+ 1) = 5x
x
2
+ 1 = 5x/2
On dividing by x on both sides, we have
(x
2
+ 1)/x = 5/2
⇒ (x + 1/x) = 5/2 … (1)
Now, consider the expansion of (x + 1/x)
2
(x + 1/x)
2
= x
2
+ 1/x
2
+ 2
(5/2)
2
= x
2
+ 1/x
2
+ 2 … [From (1)]
x
2
+ 1/x
2
= 25/4 – 2
= (25 – 8)/4
= 17/4 … (2)
Now,
(x – 1/x)
2
= x
2
+ 1/x
2
– 2
= 17/4 – 2 … [From (2)]
= (17 – 8)/4
= 9/4
So,
x – 1/x = √9/4
Thus,
(i) x – 1/x = ±3/2 … (3)
Next, we know that
(x
3
– 1/x
3
) = (x – 1/x)
3
+ 3(x – 1/x)
= (±3/2)
3
+ 3(±3/2) … [From (3)]
= ± 27/8 ± 9/2
= ± (27 + 36)/8
= ± 63/8
(ii) Thus, x
3
– 1/x
3
= ±63/8
6. If a
2
+ b
2
= 34 and ab = 12; find:
(i) 3(a + b)
2
+ 5(a – b)
2
(ii) 7(a – b)
2
– 2(a + b)
2
Solution:
We have, a
2
+ b
2
= 34 and ab= 12
We know that,
(a + b)
2
= (a
2
+ b
2
) + 2ab
= 34 + 2 x 12
= 34 + 24
= 58
Also, we know that
(a – b)
2
= (a
2
+ b
2
) – 2ab
= 34 – 2 x 12
= 34- 24
= 10
(i) 3(a + b)
2
+ 5(a – b)
2
= 3 x 58 + 5 x 10
= 174 + 50
= 224
(ii) 7(a – b)
2
– 2(a + b)
2
= 7 x 10 – 2 x 58
= 70 – 116
= -46
7. If 3x – 4/x = 4 and x ≠ 0; find: 27x
3
– 64/x
3
.
Solution:
Given, 3x – 4/x = 4
Now, let’s consider the expansion of (3x – 4/x)
3
(3x – 4/x)
3
= 27x
3
– 64/x
3
– 3 × 3x × 4/x(3x – 4/x)
(4)
3
= 27x
3
– 64/x
3
– 36(3x – 4/x)
64 = 27x
3
– 64/x
3
– 36(4)
64 = 27x
3
– 64/x
3
– 144
27x
3
– 64/x
3
= 144 + 64
Hence,
27x
3
– 64/x
3
= 208
8. If x
2
+ 1/x
2
= 7 and x ≠ 0; find the value of: 7x
3
+ 8x – 7/x
3
– 8/x.
Solution:
Given, x
2
+ 1/x
2
= 7
On subtracting 2 from both sides, we get
x
2
+ 1/x
2
– 2 = 7 – 2
(x – 1/x)
2
= 5
x – 1/x = ±√5 … (1)
Now, consider
(x – 1/x)
3
= x
3
– 1/x
3
– 3(x – 1/x)
(±√5)
3
= x
3
– 1/x
3
– 3(±√5)
x
3
– 1/x
3
= (±√5)
3
+ 3(±√5) … (2)
Taking,
7x
3
+ 8x – 7/x
3
– 8/x
= 7x
3
– 7/x
3
+ 8x – 8/x
= 7(x
3
– 1/x
3
) + 8(x – 1/x)
= 7[(±√5)
3
+ 3(±√5)] + 8(±√5)
= ±35√5
± 21√5
± 8√5
= ±64√5
9. If x = 1/(x – 5) and x ≠ 5, find x
2
– 1/x
2
.
Solution:
Given, x = 1/(x – 5)
By cross multiplying, we have
x (x – 5) = 1
x
2
– 5x = 1
x
2
– 1 = 5x
Dividing both sides by x,
(x
2
– 1)/x = 5
(x – 1/x) = 5 … (1)
Now,
(x – 1/x)
2
= 5
2
x
2
+ 1/x
2
– 2 = 25
x
2
+ 1/x
2
= 25 + 2
= 27 … (2)
Considering the expansion (x + 1/x)
2
(x + 1/x)
2
= x
2
+ 1/x
2
+ 2
(x + 1/x)
2
= 27 + 2 … [From (1)]
(x + 1/x)
2
= 29
x + 1/x = ±√29 … (3)
We know that,
x
2
– 1/x
2
= (x + 1/x) (x – 1/x)
= (±√29) (5) … [From (3)]
= ±5√29
10. If x = 1/(5 – x) and x ≠ 5; find x
3
+ 1/x
3
.
Solution:
Given, x = 1/(5 – x)
By cross multiplying, we have
x (5 – x) = 1
x
2
– 5x = -1
x
2
+ 1 = 5x
Dividing both sides by x,
(x
2
+ 1)/x = 5
x + 1/x = 5 … (1)
Now,
(x + 1/x)
3
= x
3
+ 1/x
3
+ 3(x + 1/x)
x
3
+ 1/x
3
= (x + 1/x)
3
– 3(x + 1/x)
= 5
3
– 3(5)
= 125 – 15
= 110
Thus, x
3
+ 1/x
3
= 110
11. If 3a + 5b + 4c = 0,
Show that: 27a
3
+ 125b
3
+ 64c
3
= 180abc
Solution:
Given, 3a + 5b + 4c = 0
⇒ 3a + 5b = -4c
On cubing on both sides, we have
(3a + 5b)
3
= (-4c)
3
(3a)
3
+ (5b)
3
+ 3 x 3a x 5b (3a + 5b) = -64c
3
27a
3
+ 125b
3
+ 45ab (-4c) = -64c
3
27a
3
+ 125b
3
– 180abc = -64c
3
27a
3
+ 125b
3
+ 64c
3
= 180abc
– Hence Proved.
12. The sum of two numbers is 7 and the sum of their cubes is 133, find the sum of their square.
Solution:
Let’s assume a and b to be the two numbers
So, a + b = 7 and a
3
+ b
3
= 133
We know that,
(a + b)
3
= a
3
+ b
3
+ 3ab (a + b)
(7)
3
= 133 + 3ab (7)
343 = 133 + 21ab
21ab = 343 – 133
= 210
⇒ ab = 21
Now,
a
2
+ b
2
= (a + b)
2
– 2ab
= 7
2
– 2 x 10
= 49 – 20
= 29
13. In each of the following, find the value of ‘a’:
(i) 4x
2
+ ax + 9 = (2x + 3)
2
(ii) 4x
2
+ ax + 9 = (2x – 3)
2
(iii) 9x
2
+ (7a – 5)x + 25 = (3x + 5)
2
Solution:
(i) 4x
2
+ ax + 9 = (2x + 3)
2
= 4x
2
+ 12x + 9
On comparing coefficients of x terms, we get
ax = 12x
So,
a = 12
(ii) 4x
2
+ ax + 9 = (2x – 3)
2
= 4x
2
+ 12x + 9
On comparing coefficients of x terms, we get
ax = -12x
So,
a = -12
(iii) 9x
2
+ (7a – 5)x + 25 = (3x + 5)
2
= 9x
2
+ 30x + 25
On comparing coefficients of x terms, we get
(7a – 5)x = 30x
7a – 5 = 30
7a = 35
⇒ a = 5
14. If (x
2
+ 1)/x = 3 1/3 and x > 1; find
(i) x – 1/x
(ii) x
3
– 1/x
3
Solution:
Given,
(x
2
+ 1)/x = 3 1/3 = 10/3
x + 1/x = 10/3
On squaring on both sides, we get
(x + 1/x)
2
= (10/3)
2
x
2
+ 1/x
2
+ 2 = 100/9
x
2
+ 1/x
2
= 100/9 – 2
= (100 – 18)/9
= 82/9
Now,
(x – 1/x)
2
= x
2
+ 1/x
2
– 2
= 82/9 – 2
= (82 – 18)/9
= 64/9
x – 1/x = √(64/9)
= ±8/3
On cubing both sides, we get
(x – 1/x)
3
= (8/3)
3
x
3
– 1/x
3
– 3(x – 1/x) = 512/27
x
3
– 1/x
3
= 3(x – 1/x) + 512/27
= 3(8/3) + 512/27
= 24/3 + 512/27
= (216 + 512)/27
= 728/27
Therefore, x
3
– 1/x
3
= 728/27
15. The difference between two positive numbers is 4 and the difference between their cubes is 316.
Find:
(i) Their product
(ii) The sum of their squares
Solution:
Given, difference between two positive numbers is 4
And, the difference between their cubes is 316
Let’s assume the positive numbers to be a and b
So,
a – b = 4
a
3
– b
3
= 316
On cubing both sides, we have
(a – b)
3
= 64
a
3
– b
3
– 3ab(a – b) = 64
Also,
Given: a
3
– b
3
= 316
So,
316 – 64 = 3ab(4)
252 = 12ab
So,
ab = 21
Thus, the product of numbers is 21
Now,
On squaring both sides, we get
(a – b)
2
= 16
a
2
+ b
2
– 2ab = 16
a
2
+ b
2
= 16 + 42 = 58
Thus, sum of their squares is 58.
Exercise 4(E)
1. Simplify:
(i) (x + 6)(x + 4)(x – 2)
(ii) (x – 6)(x – 4)(x + 2)
(iii) (x – 6)(x – 4)(x – 2)
(iv) (x + 6)(x – 4)(x – 2)
Solution:
Using identity:
(x + a)(x + b)(x + c) = x
3
+ (a + b + c)x
2
+ (ab + bc + ca)x + abc
(i) We have, (x + 6)(x + 4)(x – 2)
= x
3
+ (6 + 4 – 2)x
2
+ [6 × 4 + 4 × (-2) + (-2) × 6]x + 6 × 4 × (-2)
= x
3
+ 8x
2
+ (24 – 8 – 12)x – 48
= x
3
+ 8x
2
+ 4x – 48
(ii) We have, (x – 6)(x – 4)(x + 2)
= x
3
+ (-6 – 4 + 2)x
2
+ [-6 × (-4) + (-4) × 2 + 2 × (-6)]x + (-6) × (-4) × 2
= x
3
– 8x
2
+ (24 – 8 – 12)x + 48
= x
3
– 8x
2
+ 4x + 48
(iii) We have, (x – 6)(x – 4)(x – 2)
= x
3
+ (-6 – 4 – 2)x
2
+ [-6 × (-4) + (-4) × (-2) + (-2) × (-6)]x + (-6) × (-4) × (-2)
= x
3
– 12x
2
+ (24 + 8 + 12)x – 48
= x
3
– 12x
2
+ 44x – 48
(iv) We have, (x + 6)(x – 4)(x – 2)
= x
3
+ (6 – 4 – 2)x
2
+ [6 × (-4) + (-4) × (-2) + (-2) × 6]x + 6 × (-4) × (-2)
= x
3
– 0x
2
+ (-24 + 8 – 12)x + 48
= x
3
– 28x + 48
2. Simply using following identity:
(a ± b) (a
2
∓ ab + b
2
) = a
3
± b
3
(i) (2x + 3y) (4x
2
– 6xy + 9y
2
)
(ii) (3x – 5/x) (9x
2
+ 15 + 25/x
2
)
(iii) (a/3 – 3b) (a
2
+ ab + 9b
2
)
Solution:
(i) We have, (2x + 3y) (4x
2
– 6xy + 9y
2
)
= (2x + 3y) [(2x)
2
– (2x)(3y) + (3y)
2
]
= (2x)
3
+ (3y)
3
= 8x
3
+ 27y
3
(ii) We have, (3x – 5/x) (9x
2
+ 15 + 25/x
2
)
= (3x – 5/x) [(3x)
2
+ (3x)(5/x) + (5/x)
2
]
= (3x)
3
+ (5/x)
3
= 27x
3
+ 125/x
3
(iii) We have, (a/3 – 3b) (a
2
/9 + ab + 9b
2
)
= (a/3 – 3b) [(a/3)
2
+ (a/3)(3b) + (3b)
2
]
= (a/3)
3
– (3b)
3
= a
3
/27 – 27b
3
3. Using suitable identity, evaluate
(i) (104)
3
(ii) (97)
3
Solution:
Using identity: (a ± b)
3
= a
3
± b
3
± 3ab(a ± b)
(i) (104)
3
= (100 + 4)
3
= (100)
3
+ (4)
3
+ 3 × 100 × 4(100 + 4)
= 1000000 + 64 + 1200 × 104
= 1000000 + 64 + 124800
= 1124864
(ii) (97)
3
= (100 – 3)
3
= (100)
3
– (3)
3
– 3 × 100 × 3(100 – 3)
= 1000000 – 27 – 900 × 97
= 1000000 – 27 – 87300
= 912673
4. Simply:
Solution:
We know that,
If a + b + c = 0, then a
3
+ b
3
+ c
3
= 3abc
Now, if
(x
2
– y
2
) + (y
2
– z
2
) + (z
2
– x
2
) = 0
Then, we have
(x
2
– y
2
)
3
+ (y
2
– z
2
)
3
+ (z
2
– x
2
)
3
= 3(x
2
– y
2
)(y
2
– z
2
)(z
2
– x
2
) … (1)
Similarly, if
x – y + y – z + z – x = 0
Then,
(x – y)
3
+ (y – z)
3
+ (z – x)
3
= 3(x – y)(y – z)(z – x) … (2)
Now,
= (x + y)(y + z)(z + x)
5. Evaluate:
Solution:
(i) We have,
= a + b
= 0.8 + 0.5
= 1.3
(ii) We have,
6. If a – 2b + 3c = 0; state the value of a
3
– 8b
3
+ 27c
3
.
Solution:
Given, a – 2b + 3c = 0
Then,
a
3
– 8b
3
+ 27c
3
= a
3
+ (-2b)
3
+ (3c)
3
= 3(a)( -2b)(3c)
= -18abc
7. If x + 5y = 10; find the value of x
3
+ 125y
3
+ 150xy – 1000.
Solution:
Given, x + 5y = 10
On cubing both sides, we get
(x + 5y)
3
= 10
3
x
3
+ (5y)
3
+ 3(x)(5y)(x + 5y) = 1000
x
3
+ (5y)
3
+ 3(x)(5y)(10) = 1000
x
3
+ (5y)
3
+ 150xy = 1000
Thus,
x
3
+ (5y)
3
+ 150xy – 1000 = 0
8. If x = 3 + 2√2, find:
(i) 1/x
(ii) x – 1/x
(iii) (x – 1/x)
3
(iv) x
3
– 1/x
3
Solution:
We have, x = 3 + 2√2
(i) 1/x = 1/(3 + 2√2)
= (3 – 2√2)/ [(3 + 2√2) × (3 – 2√2)]
= (3 – 2√2)/ [3
2
– (2√2)
2
]
= (3 – 2√2)/ (9 – 8)
= 3 – 2√2
(ii) x – 1/x = (3 + 2√2) – (3 – 2√2) … [From (i)]
= (3 + 2√2 – 3 + 2√2)
= 4√2
(iii) (x – 1/x)
3
= (4√2)
3
… [From (ii)]
= (64 x 2√2)
= 128√2
(iv) (x
3
– 1/x
3
) = (x – 1/x)
3
– 3(x – 1/x)
= 128√2 – 3(4√2) … [From (iii) and (ii)]
= 128√2 – 12√2
9. If a + b = 11 and a
2
+ b
2
= 65; find a
3
+ b
3
.
Solution:
Given, a + b = 11 and a
2
+ b
2
= 65
Now, we know that
(a + b)
2
= a
2
+ b
2
+ 2ab
(11)
2
= 65 + 2ab
121 = 65 + 2ab
2ab = 121 – 65
ab = (121 – 65)/2
= 56/2
= 28
Considering the expansion (a
3
+ b
3
)
(a
3
+ b
3
) = (a + b) (a
2
+ b
2
– ab)
= (11) (65 – 28)
= 11 × 37
= 407
Thus, a
3
+ b
3
= 407
10. Prove that:
x
2
+ y
2
+ z
2
– xy – yz – zx is always positive.
Solution:
We have, x
2
+ y
2
+ z
2
– xy – yz – zx
= 2(x
2
+ y
2
+ z
2
– xy – yz – zx)
= 2x
2
+ 2y
2
+ 2z
2
– 2xy – 2yz – 2zx
= x
2
+ x
2
+ y
2
+ y
2
+ z
2
+ z
2
– 2xy – 2yz – 2zx
= (x
2
+ y
2
– 2xy) + (z
2
+ x
2
– 2zx) + (y
2
+ z
2
– 2yz)
= (x – y)
2
+ (z – x)
2
+ (y – z)
2
As the square of any number is positive, the given equation is always positive.
11. Find:
(i) (a + b)(a + b)
(ii) (a + b)(a + b)(a + b)
(iii) (a – b)(a – b)(a – b) by using the result of part (ii)
Solution:
(i) We have, (a + b)(a + b)
= (a + b)
2
= a × a + a × b + b × a + b × b
= a
2
+ ab + ab + b
2
= a
2
+ b
2
+ 2ab
(ii) We have, (a + b)(a + b)(a + b)
= (a × a + a × b + b × a + b × b)(a + b)
= (a
2
+ ab + ab + b
2
)(a + b)
= (a
2
+ b
2
+ 2ab)(a + b)
= a
2
× a + a
2
× b + b
2
× a + b
2
× b + 2ab × a + 2ab × b
= a
3
+ a
2
b + ab
2
+ b
3
+ 2a
2
b + 2ab
2
= a
3
+ b
3
+ 3a
2
b + 3ab
2
(iii) We have, (a – b)(a – b)(a – b)
In result (ii), replacing b by -b, we get (a – b)(a – b)(a – b)
= a
3
+ (-b)
3
+ 3a
2
(-b) + 3a(-b)
2
= a
3
– b
3
– 3a
2
b + 3ab
2
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