NCERT Solutions Class 9 Mathematics

Exercise 2.5

Q.1. Use suitable identities to find the following products:

(i) (x + 4) (x + 10) (ii) (x + 8) (x - 10)

(iii) (3x + 4) (3x - 5) (iv) (y2+ $\frac{3}{2}$) (y2 - $\frac{3}{2}$)

(v) (3 - 2x) (3 + 2x)

Q.2. Evaluate the following products without multiplying directly:

(i) 103 × 107 (ii) 95 × 96 (iii) 104 × 96

Q.3. Factorise the following using appropriate identities:

(i) 9x2 + 6xy + y(ii) 4y2 - 4y + 1  (iii) x- $\frac{{\mathrm{y}}^{2}}{100}$

Q.4. Expand each of the following, using suitable identities:

 (i) (x + 2y + 4z)2 (ii) (2x - y + z)2 (iii) (-2x + 3y + 2z)2 (iv) (3a - 7b - c)2 (v) (-2x + 5y - 3z)2 (vi) [$\frac{1}{4}$ a - $\frac{1}{2}$ b + 1]2

Q.5. Factorise:

(i) 4x2 + 9y2 + 16z2 + 12xy - 24yz - 16xz

(ii) 2x2 + y2 + 8z2 - 2√2 xy + 4√2 yz - 8xz

Q.6. Write the following cubes in expanded form:

(i) (2x + 1)3 (ii) (2a - 3b)3 (iii) [x + 1]3 (iv) [x - $\frac{2}{3}$ y]3

Q.7. Evaluate the following using suitable identities:

(i) (99)3 (ii) (102)3 (iii) (998)3

Q.8. Factorise each of the following:

(i) 8a3 + b3 + 12a2b + 6ab2

(ii) 8a3 - b3 - 12a2b + 6ab2

(iii) 27 - 125a3 - 135a + 225a2

(iv) 64a3 - 27b3 - 144a2b + 108ab2

Q.9. Verify:

(i) x3 + y3 = (x + y) (x2 - xy + y2)

(ii) x3 - y3 = (x - y) (x2 + xy + y2)

Q.10. Factorise each of the following:

(i) 27y3 + 125z3 (ii) 64m3 - 343n3

Q.11. Factorise: 27x3 + y3 + z3 - 9xyz

Q.12. Verify that: x3 + y3 + z3 - 3xyz

= $\frac{1}{2}$(x + y + z) [(x - y)+ (y - z)+ (z - x)2]

Q.13. If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.

Q.14. Without actually calculating the cubes, find the value of each of the following:

(i) (-12)3 + (7)3 + (5)3 (ii) (28)3 + (-15)3 + (-13)3

Q.15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:

(i) Area: 25a2 - 35a + 12 (ii) Area: 35 y2 + 13y - 12

Q.16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?

(i) Volume: 3x2 - 12x (ii) Volume: 12ky2 + 8ky - 20k

NCERT Solutions Class 9 Mathematics

Exercise 2.5

Q.1. Use suitable identities to find the following products:

(i) (x + 4) (x + 10) (ii) (x + 8) (x - 10)

(iii) (3x + 4) (3x - 5) (iv) (y2+ $\frac{3}{2}$) (y2 - $\frac{3}{2}$)

(v) (3 - 2x) (3 + 2x)

Solution:

(i) Using identity, (x + a)(x + b) = x2 + (a + b)x + ab

(x + 4) (x + 10) = x2 + (4 + 10)x + (4 × 10)

= x2 + 14x + 40

(ii) Using identity, (x + a)(x + b) = x2 +(a + b)x + ab

(x + 8) (x -10) = x2 + {8 +(- 10)}x + {8×(-10)}

= x2 + (8 - 10)x - 80

= x2 - 2x - 80

(iii) Using identity, (x + a)(x +b) = x2 +(a + b)x + ab

(3x + 4) (3x - 5)

= (3x)2 + {4 + (-5)}3x + {4×(-5)}

= 9x2 + 3x(4 - 5) - 20

= 9x2 - 3x - 20

(iv) Using identity, (x + y) (x -y) = x2 - y2

= y4 - $\frac{9}{4}$

(v) Using identity, (x + y) (x -y) = x2 - y2

(3 - 2x) (3 + 2x) = 32 - (2x)2

= 9 - 4x2

Q.2. Evaluate the following products without multiplying directly:

(i) 103 × 107 (ii) 95 × 96 (iii) 104 × 96

Solution:

1. Using identity, (x + a) (x + b) = x2 + (a + b) x + ab, we can write,

103 × 107 = (100 + 3) (100 + 7)

= (100)2 + (3 + 7)10 + (3 × 7)

= 10000 + 100 + 21

= 10121

2. Using identity, (x + a) (x + b) = x2 + (a + b) x + ab, we can write,

95 × 96 = (90 + 5) (90 + 4)

= 902 + 90(5 + 6) + (5 × 6)

= 8100 + (11 × 90) + 30

= 8100 + 990 + 30 = 9120

3. Using identity, (x + y) (x -y) = x2 - y2, we can write,

104 × 96 = (100 + 4) (100 - 4)

= (100)2 - (4)

= 10000 - 16 = 9984

Q.3. Factorise the following using appropriate identities:

(i) 9x2 + 6xy + y(ii) 4y2 - 4y + 1  (iii) x- $\frac{{\mathrm{y}}^{2}}{100}$

Solution:

(i) Using identity, (a + b)2 = a2 + 2ab + b2

9x2 + 6xy + y= (3x) 2 + (2×3x×y) + y2

= (3x + y)= (3x + y) (3x + y)

(ii) Using identity, (a - b)2 = a2 - 2ab + b2

4y2 - 4y + 1 = (2y)2 - (2×2y×1) + 12

= (2y - 1)= (2y - 1) (2y - 1)

(iii) Using identity, a2 - b2 = (a + b) (a - b)

x- $\frac{{\mathrm{y}}^{2}}{100}$ = x- ${\left(\frac{\mathrm{y}}{10}\right)}^{2}$

= (x - $\frac{\mathrm{y}}{10}$) (x + $\frac{\mathrm{y}}{10}$)

Q.4. Expand each of the following, using suitable identities:

 (i) (x + 2y + 4z)2 (ii) (2x - y + z)2 (iii) (-2x + 3y + 2z)2 (iv) (3a - 7b - c)2 (v) (-2x + 5y - 3z)2 (vi) [$\frac{1}{4}$ a - $\frac{1}{2}$ b + 1]2

Solution:

(i) Using identity, (a + b + c)= a2 + b2 + c2 + 2ab + 2bc + 2ca

We can write,

(x + 2y + 4z)2 = x2 + (2y)2 + (4z)2 + (2×x×2y)+(2×2y×4z)+(2×4z×x)

= x2 + 4y2 + 16z2 + 4xy + 16yz + 8xz

(ii) Using identity, (a + b + c)= a2 + b2 + c2 + 2ab + 2bc + 2ca

We can write,

(2x - y + z)= (2x)2 + (-y)2 + z2 + (2×2x×-y) +(2×-y×z) + (2×z×2x)

= 4x2 + y2 + z2 - 4xy - 2yz + 4xz

(iii) Using identity, (a + b + c)= a2 + b2 + c2 + 2ab + 2bc + 2ca

We can write,

(-2x + 3y + 2z)2

= (-2x)2 + (3y)2 + (2z)2 + (2×-2x×3y) +(2×3y×2z)+ (2×2z×-2x)

= 4x2 + 9y2 + 4z2 - 12xy + 12yz - 8xz

(iv)Using identity, (x + y + z)= x2 + y2 + z2 + 2xy + 2yz + 2zx

We can write,

(3a - 7b - c)2

= (3a)2 + (-7b)2 + (-c)2 + (2×3a×-7b) + (2×-7b×-c) + (2×-c×3a)

= 9a2 + 49b2 + c2 - 42ab + 14bc - 6ac

(v) Using identity, (a + b + c)= a2 + b2 + c2 + 2ab + 2bc + 2ca

We can write,

(-2x + 5y - 3z)2

= (-2x)2 + (5y)2 + (-3z)2 +(2×-2x×5y)+(2×5y×-3z)+(2×-3z×-2x)

= 4x2 + 25y2 + 9z2 - 20xy - 30yz + 12xz

(vi)Using identity, (a + b + c)= a2 + b2 + c2 + 2ab + 2bc + 2ca

We can write,

$+\left(2×\frac{1}{4}\mathrm{a}×-\frac{1}{2}\mathrm{b}\right)+\left(2×-\frac{1}{2}\mathrm{b}×1\right)+\left(2×1×\frac{1}{4}\mathrm{a}\right)$

Q.5. Factorise:

(i) 4x2 + 9y2 + 16z2 + 12xy - 24yz - 16xz

(ii) 2x2 + y2 + 8z2 - 2√2 xy + 4√2 yz - 8xz

Solution:

(i) Using identity, (a + b + c)= a2 + b2 + c2 + 2ab + 2bc + 2ca

We can write,

4x2 + 9y2 + 16z2 + 12xy - 24yz - 16xz

= (2x)2 + (3y)2 + (-4z)2 + (2×2x×3y) + (2×3y×-4z) + (2×-4z×2x)

= (2x + 3y - 4z)2

= (2x + 3y - 4z) (2x + 3y - 4z)

(ii) Using identity, (a + b + c)= a2 + b2 + c2 + 2ab + 2bc + 2ca

We can write,

2x2 + y2 + 8z2 - 2$\sqrt{2}$ xy + 4$\sqrt{2}$ yz - 8xz

= (- $\sqrt{2}$x)2 + (y)2 + (2$\sqrt{2}$z)2 + 2×(-$\sqrt{2}$x) ×y + 2×y×2$\sqrt{2}$z + 2×2$\sqrt{2}$z×(-$\sqrt{2}$x)

= (-$\sqrt{2}$x + y + 2$\sqrt{2}$z)2

= (-$\sqrt{2}$x + y + 2$\sqrt{2}$z) (-$\sqrt{2}$x + y + 2$\sqrt{2}$z)

Q.6. Write the following cubes in expanded form:

(i) (2x + 1)3 (ii) (2a - 3b)3 (iii) [x + 1]3 (iv) [x - $\frac{2}{3}$ y]3

Solution:

(i) Using identity, (a + b)3 = a3 + b3 + 3ab(a + b)

We can write,

(2x + 1)= (2x)3 + 13 + (3×2x×1)(2x + 1)

= 8x3 + 1 + 6x(2x + 1)

= 8x3 + 12x2 + 6x + 1

(ii) Using identity, (x - y)3 = x3 - y3 - 3xy(x - y)

We can write,

(2a - 3b)3 = (2a)3 - (3b)3 - (3×2a×3b)(2a - 3b)

= 8a3 - 27b3 - 18ab(2a - 3b)

= 8a3 - 27b3 - 36a2b + 54ab2

(iii) Using identity, (a + b)3 = a3 + b3 + 3ab(a + b)

We can write,

(iv) Using identity, (a - b)3 = a3 - b3 - 3ab(a -b)

We can write,

Q.7. Evaluate the following using suitable identities:

(i) (99)3 (ii) (102)3 (iii) (998)3

Solution:

(i) We can write, (99)3 = (100 - 1)3

Using identity, (a - b)3 = a3 - b3 - 3ab(a - b),

(100 - 1)= (100)3 - 13 - (3×100×1)(100 - 1)

= 1000000 - 1 - 300(100 - 1)

= 1000000 - 1 - 30000 + 300

= 970299

(ii) We can write, (102)3 = (100 + 2)3

Using identity, (a + b)3 = a3 + b3 + 3ab(a + b),

(100 + 2)= (100)3 + 23 + (3×100×2)(100 + 2)

= 1000000 + 8 + 600(100 + 2)

= 1000000 + 8 + 60000 + 1200

= 1061208

(iii)We can write, (998)3 = (1000 - 2)

Using identity, (a - b)3 = a3 - b3 - 3ab(a - b)

(1000-2)= (1000)3 - 23 - (3×1000×2)(1000 - 2)

= 100000000 - 8 - 6000(1000 - 2)

= 100000000 - 8- 600000 + 12000

= 994011992

Q.8. Factorise each of the following:

(i) 8a3 + b3 + 12a2b + 6ab2

(ii) 8a3 - b3 - 12a2b + 6ab2

(iii) 27 - 125a3 - 135a + 225a2

(iv) 64a3 - 27b3 - 144a2b + 108ab2

Solution:

(i) Using identity, (a + b)3 = a3 + b3 + 3a2b + 3ab2

8a3 + b3 + 12a2b + 6ab2

= (2a)3 + b3 + 3(2a)2b + 3(2a)(b)2

= (2a + b)3

= (2a + b)(2a + b)(2a + b)

(ii) Using identity, (a - b)3 = a3 - b3 - 3a2b + 3ab2

8a3 - b3 - 12a2b + 6ab2

= (2a)3 - b3 - 3(2a)2b + 3(2a)(b)2

= (2a - b)3

= (2a - b)(2a - b)(2a - b)

(iii) Using identity, (a - b)3 = a3 - b3 - 3a2b + 3ab2

27 - 125a3 - 135a + 225a2

= 33 - (5a)3 - 3(3)2(5a) + 3(3)(5a)2

= (3 - 5a)3

= (3 - 5a)(3 - 5a)(3 - 5a)

(iv) Using identity, (a - b)3 = a3 - b3 - 3a2b + 3ab2

64a3 - 27b3 - 144a2b + 108ab2

= (4a)3 - (3b)3 - 3(4a)2(3b) + 3(4a)(3b)2

= (4a - 3b)3

= (4a - 3b)(4a - 3b)(4a - 3b)

(v) Using identity, (a - b)3 = a3 - b3 - 3a2b + 3ab2

Q.9. Verify:

(i) x3 + y3 = (x + y) (x2 - xy + y2)

(ii) x3 - y3 = (x - y) (x2 + xy + y2)

Solution:

(i) We know that,

(x + y)3 = x3 + y3 + 3xy(x + y)

⇒ x3 + y3 = (x + y)- 3xy(x + y)

= (x + y)[(x + y)2 - 3xy]

= (x + y)[(x2 + y+ 2xy) - 3xy]

Thus, x3 + y3 = (x + y)(x2 + y2 - xy)

(ii) We know that,

(x - y)3 = x3 - y3 - 3xy(x - y)

⇒ x3 - y3 = (x - y)3 + 3xy(x - y)

= (x - y)[(x - y)2 + 3xy]

= (x - y)[(x2 + y2 - 2xy) + 3xy]

Thus, x3 + y3 = (x + y)(x2 + y2 + xy)

Q.10. Factorise each of the following:

(i) 27y3 + 125z3 (ii) 64m3 - 343n3

Solution:

(i) Using identity, x3 + y3 = (x + y) (x2 - xy + y2)

We have, 27y3 + 125z3 = (3y)3 + (5z)3

= (3y + 5z) {(3y)2 - (3y)(5z) + (5z)2}

= (3y + 5z) (9y2 - 15yz + 25z)2

(ii) Using identity, x3 - y3 = (x - y) (x2 + xy + y2 )

We have, 64m3 - 343n3 = (4m)3 - (7n)3

= (4m + 7n) {(4m)2 + (4m)(7n) + (7n)2}

= (4m + 7n) (16m2 + 28mn + 49n2)

Q.11. Factorise: 27x3 + y3 + z3 - 9xyz

Solution:

Using identity,

x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - xz)

We have, 27x3 + y3 + z3 - 9xyz

= (3x + y + z) {(3x)2 + y2 + z2 - 3xy - yz - 3xz}

= (3x + y + z) (9x2 + y2 + z2 - 3xy - yz - 3xz)

Q.12. Verify that: x3 + y3 + z3 - 3xyz

= $\frac{1}{2}$(x + y + z) [(x - y)+ (y - z)+ (z - x)2]

Solution:

We know that,

x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - xz)

Now, RHS = $\frac{1}{2}$×(x + y + z) 2(x2 + y2 + z2 - xy - yz - xz)

= $\frac{1}{2}$ (x + y + z) (2x2 + 2y2 + 2z2 - 2xy - 2yz - 2xz)

= $\frac{1}{2}$ (x + y + z) [(x2 + y2 -2xy) + (y+ z2 - 2yz) + (x2 + z2 - 2xz)]

= $\frac{1}{2}$ (x + y + z) [(x - y)+ (y - z)+ (z - x)2]

Therefore,

x3 + y3 + z3 - 3xyz = $\frac{1}{2}$(x + y + z) [(x - y)+ (y - z)+ (z - x)2]

Q.13. If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.

Solution:

We know that,

x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - xz)

Now taking (x + y + z) = 0,  we get,

x3 + y3 + z3 - 3xyz = (0)(x2 + y2 + z2 - xy - yz - xz)  = 0

Q.14. Without actually calculating the cubes, find the value of each of the following:

(i) (-12)3 + (7)3 + (5)3 (ii) (28)3 + (-15)3 + (-13)3

Solution:

(i) (-12)3 + (7)3 + (5)3

Let x = -12, y = 7 and z = 5

We see that, x + y + z = -12 + 7 + 5 = 0

We know that if, x + y + z = 0, then x3 + y3 + z3 = 3xyz

⇒ (-12)3 + (7)3 + (5)3 = 3(-12)(7)(5) = -1260

(ii) (28)3 + (-15)3 + (-13)3

Let x = 28, y = -15 and z = -13

We see that, x + y + z = 28 - 15 - 13 = 0

We know that if, x + y + z = 0, then x3 + y3 + z3 = 3xyz

⇒ (28)3 + (-15)3 + (-13)3 = 3(28)(-15)(-13) = 16380

Q.15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:

(i) Area: 25a2 - 35a + 12 (ii) Area: 35 y2 + 13y - 12

Solution:

Since, area is product of length and breadth therefore by factorizing the given area, we can know the length and breadth of rectangle.

(i) Area: 25a2 - 35a + 12

We can write, 25a2 - 35a + 12 = 25a2 - 15a -20a + 12

= 5a(5a - 3) - 4(5a - 3)

= (5a - 4)(5a - 3)

Hence, possible expression for length = 5a - 3

And possible expression for breadth = 5a - 4

(ii) Area: 35 y2 + 13y - 12

We can write, 35 y2 + 13y - 12 = 35y2 - 15y + 28y - 12

= 5y(7y - 3) + 4(7y - 3)

= (5y + 4)(7y - 3)

Hence, possible expression for length = 5y + 4

And possible expression for breadth = 7y - 3

Q.16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?

(i) Volume: 3x2 - 12x (ii) Volume: 12ky2 + 8ky - 20k

Solution:

Since, volume is product of length, breadth and height therefore by factorizing the given volume, we can know the length, breadth and height of the cuboid.

(i) Volume: 3x2 - 12x

We can write, 3x2 - 12x = 3x(x - 4)

Hence, possible expression for length = 3

Possible expression for breadth = x

Possible expression for height = (x - 4)

(ii) Volume: 12ky2 + 8ky - 20k

We can write, 12ky2 + 8ky - 20k = 4k(3y2 + 2y - 5)

= 4k(3y2 +5y - 3y - 5)

= 4k[y(3y +5) - 1(3y + 5)]

= 4k (3y +5) (y - 1)

Hence, possible expression for length = 4k

Possible expression for breadth = 3y +5

Possible expression for height = y - 1