NCERT Solutions for Class 9 Maths Exercise 2.5
(i)
(ii)
(iii)
(iv)
(v)
Solution:
(i)
We know that.
We need to apply the above identity to find the product
Therefore, we conclude that the productis.
(ii)
We know that.
We need to apply the above identity to find the product
=
Therefore, we conclude that the productis.
(iii)
We know that.
We need to apply the above identity to find the product
Therefore, we conclude that the productis.
(iv)
We know that.
We need to apply the above identity to find the product
Therefore, we conclude that the productis.
(v)
We know that.
We need to apply the above identity to find the product
Therefore, we conclude that the productis.
Question 2. Evaluate the following products without multiplying directly :
(i)
(ii)
(iii)
Solution:
(i)
We can observe that, we can apply the identity
=10000+1000+21
=11021
Therefore, we conclude that the value of the productis.
(ii)
We can observe that, we can apply the identity
=10000-900+20
=9120
Therefore, we conclude that the value of the productis .
(iii)
We can observe that, we can apply the identitywith respect to the expression, to get
=10000-16
=9984
Therefore, we conclude that the value of the productis.
Question 3. Factorize the following using appropriate identities :
(i)
(ii)
(iii)
Solution:
(i)
We can observe that, we can apply the identity
(ii)
We can observe that, we can apply the identity
(iii)
We can observe that, we can apply the identity
Question 4. Expand each of the following, using suitable identities :
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Solution:
(i)
We know that.
We need to apply the above identity to expand the expression.
(ii) (2x-y+z)2
We know that.(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
We need to apply the above identity to expand the expression (2x-y+z)2
So comparing the above , a= 2x , b= -y and c= z
therefore (2x-y+z)2 = (2x)2 + (-y)2 + (z)2 + 2(2x)(-y) + 2(-y)z + 2z(2x)
= 4x2 + y2 + z2 - 4xy- 2yz +4zx
(iii) (-2x + 3y + 2z)2
We know that.(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
We need to apply the above identity to expand the expression.
So comparing the above , a= -2x , b= 3y and c= 2z
therefore (2x-y+z)2 = (-2x)2 + (3y)2 + (2z)2 + 2(-2x)(3y) + 2(3y)(2z)z + 2(2z)(-2x)
= 4x2 + 9y2 + 4z2 - 12xy + 12yz - 8zx
(iv) (3a - 7b - c)2
We know that.(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
We need to apply the above identity to expand the expression.
So comparing the above , x = 3a , y = -7b and z = -c
therefore (3a - 7b - c)2 = (3a)2 + (-7b)2 + (-c)2 + 2(3a)(-7b) + 2(-7b)(-c) + 2(-c)(3a)
= 9a2 + 49b2 + c2 - 42xy + 14bc - 6ca
(v)
We know that.
We need to apply the above identity to expand the expression.
(vi)
We know that.
Question 5. Factorize :
(i)
(ii)
Solution:
(i)
The expression can also be written as
We can observe that, we can apply the identitywith respect to the expression, to get
Therefore, we conclude that after factorizing the expression, we get.
(ii)
We need to factorize the expression.
The expression can also be written as
We can observe that, we can apply the identity (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca with respect to the expression, to get
Therefore, we conclude that after factorizing the expression, we get.
Question 6. Write the following cubes in expanded form :
(i)
(ii)
(iii)
(iv) (x - 2y/3)3
Solution :
(i)
We know that.
Therefore, the expansion of the expressionis .
(ii)
We know that.
(2a-3b)3 = (2a)3 + (-3b)3 + 3(2a)(-3b)(2a-3b)
= 8a3 - 27b3 -18ab(2a-3b)
= 8a3 - 27b3 -36a2b + 54ab2
Therefore, the expansion of the expression is 8a3 - 27b3 -36a2b + 54ab2
.
(iii) (3x/2 + 1)3
We know that. (a + b )3 = a3 + b3 + 3a2b +3ab2
Comparing the corresponding terms with the identity,
(3x/2 + 1)3 = (3x/2)3 + (1)3 + 3(3x/2)2(1) + 3(3x/2)(1)2
= 27x3/8 + 1 + 27x2/4 + 9x/2
Therefore, the expansion of the expression (3x/2 + 1)3 is 27x3/8 + 1 + 27x2/4 + 9x/2.
(iv)(x - 2y/3)3
We know that (a + b )3 = a3 + b3 + 3a2b +3ab2
Comparing the corresponding terms with the identity
(x - 2y/3)3 = (x)3 + (-2y/3)3 + 3(x)2(-2y/3) + 3(x)(-2y/3)2
= x3 - 8y3/27 -2x2y + 4xy2/3
Therefore, the expansion of the expression (x - 2y/3)3 is x3 - 8y3/27 -2x2y + 4xy2/3
.
Question 7. Evaluate the following using suitable identities
(i) (99)3
(ii) (102)3
(iii) (998)3
Solution:
(i) We have, 99 = (100 -1)
∴ 993 = (100 – 1)3
= (100)3 – 13 – 3(100)(1)(100 -1)
[Using (a – b)3 = a3 – b3 – 3ab (a – b)
= 1000000 – 1 – 300(100 – 1)
= 1000000 -1 – 30000 + 300
= 1000300 – 30001 = 970299
(ii) We have, 102 =100 + 2
∴ 1023 = (100 + 2)3
= (100)3 + (2)3 + 3(100)(2)(100 + 2)
[Using (a + b)3 = a3 + b3 + 3ab (a + b)]
= 1000000 + 8 + 600(100 + 2)
= 1000000 + 8 + 60000 + 1200 = 1061208
(iii) We have, 998 = 1000 – 2
∴ (998)3 = (1000-2)3
= (1000)3– (2)3 – 3(1000)(2)(1000 – 2)
[Using (a – b)3 = a3 – b3 – 3ab (a – b)]
= 1000000000 – 8 – 6000(1000 – 2)
= 1000000000 – 8 – 6000000 +12000
= 994011992
Question 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) 8a3 +b3 + 12a2b+6ab2
= (2a)3 + (b)3 + 6ab(2a + b)
= (2a)3 + (b)3 + 3(2a)(b)(2a + b)
= (2 a + b)3
[Using (a + b)3 = a3 + b3 + 3ab (a + b)]
= (2a + b)(2a + b)(2a + b)
(ii) 8a3 - b3 - 12a2b+6ab2
= (2a)3 – (b)3 – 3(2a)(b)(2a – b)
= (2a – b)3
[Using (a + b)3 = a3 + b3 + 3ab (a + b)]
= (2a – b) (2a – b) (2a – b)
(iii) 27-125a3 -135a+225a2
= (3)3 – (5a)3 – 3(3)(5a)(3 – 5a)
= (3 – 5a)3
[Using (a + b)3 = a3 + b3 + 3ab (a + b)]
= (3 – 5a) (3 – 5a) (3 – 5a)
(iv) 64a3 -27b3 -144a2b + 108ab2
= (4a)3 – (3b)3 – 3(4a)(3b)(4a – 3b)
= (4a – 3b)3
[Using a3 – b3 – 3 ab(a – b) = (a – b)3]
= (4a – 3b)(4a – 3b)(4a – 3b)
Question 9. Verify
(i) x3 + y3 = (x + y)(x2 – xy + y2)
(ii) x3 - y3 = (x + y)(x2 + xy + y2)
Solution:
(i) ∵ (x + y)3 = x3 + y3 + 3xy(x + y)
x3 + y3 = (x + y)(x2 – xy + y2)
⇒ (x + y)3 – 3(x + y)(xy) = x3 + y3
⇒ (x + y)[(x + y)2-3xy] = x3 + y3
⇒ (x + y)(x2 + y2 – xy) = x3 + y3
Hence, verified.
(ii) ∵ (x – y)3 = x3 – y3 – 3xy(x – y)
⇒ (x – y)3 + 3xy(x – y) = x3 – y3
⇒ (x – y)[(x – y)2 + 3xy)] = x3 – y3
⇒ (x – y)(x2 + y2 + xy) = x3 – y3
Hence, verified.
Question 10. Factorise each of the following
(i) 27y3 + 125z3
(ii) 64m3 – 343n3
[Hint See question 9]
Solution:
(i) We know that
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 + 25z2)
(ii) We know that
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)
Question 11. Factorise 27x3 + y3 + z3 – 9xyz
Solution:
We have,
27x3 + y3 + z3 – 9xyz = (3x)3 + (y)3 + (z)3 – 3(3x)(y)(z)
Using the identity,
x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx)
We have, (3x)3 + (y)3 + (z)3 – 3(3x)(y)(z)
= (3x + y + z)[(3x)3 + y3 + z3 – (3x × y) – (y × 2) – (z × 3x)]
= (3x + y + z)(9x2 + y2 + z2 – 3xy – yz – 3zx)
Question 12. Verify that
x3 + y3 + z3 – 3xyz = (x + y+z)[(x-y)2 + (y – z)2 +(z – x)2]
Solution:
R.H.S
= (x + y + z[(x-y)2 + (y – z)2 +(z – x)2]
= (x + y + 2)[(x2 + y2 – 2xy) + (y2 + z2 – 2yz) + (z2 + x2 – 2zx)]
= (x + y + 2)(x2 + y2 + y2 + z2 + z2 + x2 – 2xy – 2yz – 2zx)
= (x + y + z)[2(x2 + y2 + z2 – xy – yz – zx)]
= 2 x x (x + y + z)(x2 + y2 + z2 – xy – yz – zx)
= (x + y + z)(x2 + y2 + z2 – xy – yz – zx)
= x3 + y3 + z3 – 3xyz = L.H.S.
Hence, verified.
Question 13. If x + y + z = 0, show that x3 + y3 + z3 = 3 xyz.
Solution:
Since, x + y + z = 0
⇒ x + y = -z
⇒(x + y)3 = (-z)3
⇒ x3 + y3 + 3xy(x + y) = -z3
⇒ x3 + y3 + 3xy(-z) = -z3 [∵ x + y = -z]
⇒ x3 + y3 – 3xyz = -z3
⇒ x3 + y3 + z3 = 3xyz
Hence, if x + y + z = 0, then
x3 + y3 + z3 = 3xyz
Question 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) We have, (-12)3 + (7)3 + (5)3
Let x = -12, y = 7 and z = 5.
Then, 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)]
= 3[-420] = -1260
(ii) We have, (28)3 + (- 15)3 + (- 13)3
Let x = 28, y = -15 and z = -13.
Then, 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)
= 3(5460) = 16380
Question 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 35y2 + 13y – 12
Solution:
Area of a rectangle = (Length) x (Breadth)
(i) 25a2 – 35a + 12 = 25a2 – 20a – 15a + 12 = 5a(5a – 4) – 3(5a – 4) = (5a – 4)(5a – 3)
Thus, the possible length and breadth are (5a – 3) and (5a – 4).
(ii) 35y2+ 13y -12 = 35y2 + 28y – 15y -12
= 7y(5y + 4) – 3(5y + 4) = (5 y + 4)(7y – 3)
Thus, the possible length and breadth are (7y – 3) and (5y + 4).
Question 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:
Volume of a cuboid = (Length) x (Breadth) x (Height)
(i) We have, 3x2 – 12x = 3(x2 – 4x)
= 3 x (x – 4)
∴ The possible dimensions of the cuboid are 3, x and (x – 4).
(ii) We have, 12k2 + 8ky – 20k
= 4[3ky2 + 2ky – 5k] = 4[k(3y2 + 2y – 5)]
= 4 x k x (3y2 + 2y – 5)
= 4k[3y2 – 3y + 5y – 5]
= 4k[3y(y – 1) + 5(y – 1)]
= 4k[(3y + 5) x (y – 1)]
= 4k x (3y + 5) x (y – 1)
Thus, the possible dimensions of the cuboid are 4k, (3y + 5) and (y -1).
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