## RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities VSAQS

Question 1.

Solution:

Question 2.

Solution:

Question 3.
If a + b = 7 and ab = 12, find the value of a2 + b2.
Solution:
a + b = 7, ab = 12
Squaring both sides,
(a + b)2 = (7)2
⇒  a2 + b2 + 2ab = 49
⇒  a2 + b2 + 2 x 12 = 49
⇒ a2 + b2 + 24 = 49
⇒ a2 + b2 = 49 – 24 = 25
∴ a2 + b2 = 25

Question 4.
If a – b = 5 and ab = 12, find the value of a2 + b2 .
Solution:
a – b = 5, ab = 12
Squaring both sides,
⇒ (a – b)2 = (5)2
⇒  a2 + b2 – 2ab = 25
⇒  a2 + b2 – 2 x 12 = 25
⇒  a2 + b2 – 24 = 25
⇒  a2 + b2 = 25 + 24 = 49
∴ a2 + b2 = 49

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Hope given RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities VSAQS are helpful to complete your math homework.

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## RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS

Mark the correct alternative in each of the following:
Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.
The rationalisation factor of 2 + $$\sqrt { 3 }$$  is
(a) 2 – $$\sqrt { 3 }$$
(b) $$\sqrt { 2 }$$ + 3
(c)  $$\sqrt { 2 }$$ – 3
(d) $$\sqrt { 3 }$$ – 2
Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

Question 11.

Solution:

Question 12.

Solution:

Question 13.

Solution:

Question 14.

Solution:

Question 15.

Solution:

Question 16.

Solution:

Question 17.

Solution:

Question 18.

Solution:

Question 19.

Solution:

Question 20.

Solution:

Question 21.

Solution:

Question 22.

Solution:

Question 23.

Solution:

Question 24.

Solution:

Question 25.

Solution:

Hope given RD Sharma Class 9 Solutions Chapter 3 Rationalisation MCQS are helpful to complete your math homework.

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## RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.2

Other Exercises

Factorize each of the following expressions:
Question 1.
p3 + 27
Solution:
We know that a3 + b3 = (a + b) (a2 – ab + b2)
a3 – b3 = (a – b) (a2 + aft + b2)
p3 + 21 = (p)3 + (3)3
= (p + 3) (p2– p x 3 + 32)
= (p + 3) (p2 – 3p + 9)

Question 2.
y3 + 125
Solution:
y3 + 125 = (p)3 + (5)3
= (p + 5) (p2 – 5y + 52)
= (P + 5) (p2 – 5y + 25)

Question 3.
1 – 21a3
Solution:
1 – 21a3 = (1)3 – (3a)3
= (1 – 3a) [12 + 1 x 3a + (3a)2]
= (1 – 3a) (1 + 3a + 9a2)

Question 4.
8x3y3 + 27a3
Solution:
8x3y3 + 27a3
= (2xy + 3a) [(2xy)2 – 2xy x 3a + (3a)2]
= (2xy + 3a) (4x2y – 6xya + 9a2)

Question 5.
64a3 – b3
Solution:
64a3 – b3 = (4a)3 – (b)3
= (4a – b) [(4a)2 + 4a x b + (b)2]
= (4a – b) (16a2 + 4ab + b2)

Question 6.

Solution:

Question 7.
10x4– 10xy4
Solution:
I0x4y- 10xy4 = 10xy(x3 -y3)
= 10xy(x – y) (x2 + xy + y2)

Question 8.
54x6y + 2x3y4
Solution:
54 x6y + 2x3y4 = 2x3y(27x3 + y3)
= 2x3y[(3x)3 + (y)3]
= 2x3y(3x + y) [(3x)2 -3x x y + y2]
= 2x3y(3x + y) (9x2 -3xy + y2)

Question 9.
32a3 + 108b3
Solution:
32a3 + 108b3
= 4(8a3 + 27b3) = 4 [(2a)3 + (3 b)3]
= 4(2a + 3b) [(2a)2 – 2a x 3b + (3b)2]
= 4(2a + 3b) (4a2 – 6ab + 9b2)

Question 10.
(a – 2b)3 – 512b3
Solution:
(a – 2b)3 – 512b3
= (a – 2b)3 – (8b)3
= (a – 2b- 8b) [(a – 2b)2 + (a – 2b) x 8b + (8b)2]
= (a – 10b) [a2 + 4b2 – 4ab + 8ab – 16b2 + 64b2]
= (a – 10b) (a2 + 4ab + 52b2)

Question 11.
8x2y3 – x5
Solution:
8x2y3 – x5 = x2(8y3 – X3)
= x2(2y)3 – (x)3]
= x2[(2y – x) (2y)2 + 2y x x + (x)2]
= x2(2y – x) (4y2 + 2xy + x2)

Question 12.
1029 -3x3
Solution:
1029 – 3X3 = 3(343 – x3)                       ‘
= 3 [(7)3 – (x)3]
= 3(7 – x) (49x + 7x + x2)

Question 13.
x3y3+ 1
Solution:
x3y3 + 1 = (xy)3 + (1)3
= (xy + 1) [(xy)2 – xy x 1 + (1)2]
= (xy + 1) (x2y2 – xy + 1)
= (xy + 1) (x2y – xy + 1)

Question 14.
x4y4 – xy
Solution:
x4y4 – xy = xy(x3y3 – 1)
= xy[(xy3-(1)3]
= xy (xy – 1) [x2y2 + 2xy + 1]

Question 15.
a3 + b3 + a + b
Solution:
a3 + b3 + a + b
= (a + b) (a2 – ab + b2) + 1 (a + b)
= (a + b) (a2 – ab + b2 + 1)

Question 16.
Simplify:

Solution:

Question 17.
(a + b)3 – 8(a – b)3
Solution:
(a + b)3 – 8(a – b)3
= (a + b)2 – (2a – 2b)3
= (a+ b – 2a + 2b) [(a + b)2 + (a + b) (2a-2b) + (2a – 2b)2)]
= (3b – a) [a2 + b2 + 2ab + 2a22ab + 2ab – 2b2 + 4a2 – 8ab + 4b2]
= (3b – a) [7a2 – 6ab + 3b2]

Question 18.
(x + 2)3 + (x- 2)3
Solution:
(x + 2)3 + (x – 2)3
= (x + 2 + x – 2) [(x + 2)2 – (x + 2) (x – 2) + (x – 2)2]
= 2x [x2 + 4x + 4 – (x2 + 2x – 2x – 4) + x4x + 4]
= 2x[x2 + 4x + 4- x2-2x + 2x + 4+ x2– 4x + 4]
= 2x[x2 + 12]

Question 19.
x6 +y6
Solution:
x6 + y= (x2)3 + (y2)3
= (x2 + y2) [x4 – x2y2 + y4]

Question 20.
a12 + b12
Solution:
a12 + b12 = (a4)3 + (b4)3
= (a4 + b4) [(a4)2 – a4b4 + (b4)2]
= (a4 + b4) (a8 – a4b4 + b8)

Question 21.
x3 + 6x2 + 12x + 16
Solution:
x3 + 6x2 + 12x + 16
= (x)3 + 3.x2.2 + 3.x.4 + (2)3 + 8           {∵ a3 + 3a2b + 3ab2 +b3 = (a + b)3}
= (x + 2)3 + 8 = (x + 2)3 + (2)3
= (x + 2 + 2) [(x + 2)2 – (x + 2) x 2 + (2)2] {∵ a3 + b2 = (a + b) (a2 – ab + b2}
= (x + 4) (x2 + 4x + 4 – 2x – 4 + 4)
= (x + 4) (x2 + 2x + 4)

Question 22.

Solution:

Question 23.
a3 + 3a2b + 3ab2 + b3 – 8
Solution:
a3 + 3a2b + 3ab2 + b3 – 8
= (a + b)3 – (2)3
= (a + b -2)[(a + b)2 + (a +b)x2 + (2)2]
= (a + b-2) (a2 + b2 + 2ab + 2a + 2b + 4)
= (a + b – 2) (a2 + b2 + 2ab + 2(a + b) + 4]
= (a + b – 2) [(a + b)2 + 2(a + b) + 4}

Question 24.
8a3 – b3 – 4ax + 2bx
Solution:
8a3 – b3 – 4ax + 2bx
(2a)3 – (b)3 – 2x(2a – b)
= (2a-b)[(2a)2 + 2a x b + (b)2]- 2x(2a-b)
= (2a – b) [4a2 + 2ab + b2] – 2x(2a – b)
= (2a – b) [4a2 + 2ab + b2 – 2x]

Hope given RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.2 are helpful to complete your math homework.

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## RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.2

Question 1.
Write the following in the expanded form:

Solution:

Question 2.
If a + b + c = 0 and a2 + b2 + c2 = 16, find the value of ab + be + ca.
Solution:
a + b+ c = 0
Squaring both sides,
(a + b + c)2 = 0
⇒ a2 + b2 + c2 + 2(ab + bc + ca) = 0
16 + 2(ab + bc + c) = 0
⇒ 2(ab + bc + ca) = -16
⇒  ab + bc + ca =-$$\frac { 16 }{ 2 }$$ = -8
∴ ab + bc + ca = -8

Question 3.
If a2 + b2 + c2 = 16 and ab + bc + ca = 10, find the value of a + b + c.
Solution:
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
= 16 + 2 x 10
= 16 + 20 = 36
= (±6)2
∴ a + b + c = ±6

Question 4.
If a + b + c = 9 and ab + bc + ca = 23, find the value of a2 + b2 + c2.
Solution:
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
⇒ (9)2 = a2 + b2 + c2 + 2 x 23
⇒ 81= a2 + b2 + c2 + 46
⇒  a2 + b2 + c2 = 81 – 46 = 35
∴
a2 + b2 + c2 = 35

Question 5.
Find the value of 4x2 + y2 + 25z2 + 4xy – 10yz – 20zx when x = 4, y = 3 and z = 2.
Solution:
x = 4, y – 3, z = 2
4x2 + y2 + 25z2 + 4xy – 10yz – 20zx
= (2x)2 + (y)2 + (5z)2 + 2 x2 x x y-2 x y x 5z – 2 x 5z x 2x
= (2x + y- 5z)2
= (2 x 4 + 3- 5 x 2)2
= (8 + 3- 10)2
= (11 – 10)2
= (1)2 = 1

Question 6.
Simplify:
(i)  (a + b + c)2 + (a – b + c)2
(ii) (a + b + c)2 –  (a – b + c)2
(iii) (a + b + c)2 +   (a – b + c)2 + (a + b – c)2
(iv) (2x + p – c)2 – (2x – p + c)2
(v) (x2 + y2 – z2)2 – (x2 – y2 + z2)2
Solution:

Question 7.
Simplify each of the following expressions:

Solution:

Hope given RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.2 are helpful to complete your math homework.

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## RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.4

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.4

Question 1.
Find the following products:
(i) (3x + 2y) (9X2 – 6xy + Ay2)
(ii) (4x – 5y) (16x2 + 20xy + 25y2)
(iii) (7p4 + q) (49p8 – 7p4q + q2)

Solution:

Question 2.
If x = 3 and y = -1, find the values of each of the following using in identity:

Solution:

Question 3.
If a + b = 10 and ab = 16, find the value of a2 – ab + b2 and a2 + ab + b2.
Solution:
a + b = 10, ab = 16 Squaring,
(a + b)2 = (10)2
⇒ a2 + b2 + lab = 100
⇒ a2 + b2 + 2 x 16 = 100
⇒  a2 + b2 + 32 = 100
∴ a2 + b2 = 100 – 32 = 68
Now, a2 – ab + b2 = a2 + b2 – ab = 68 – 16 = 52
and a2 + ab + b2 = a2 + b2 + ab = 68 + 16 = 84

Question 4.
If a + b = 8 and ab = 6, find the value of a3 + b3.
Solution:
a + b = 8, ab = 6
Cubing both sides,
(a + b)3 = (8)3
⇒ a3 + b3 + 3 ab{a + b) = 512
⇒  a3 + b3 + 3 x 6 x 8 = 512
⇒  a3 + b3 + 144 = 512
⇒  a3 + b3 = 512 – 144 = 368
∴ a3 + b3 = 368

Question 5.
If a – b = 6 and ab = 20, find the value of a3-b3.
Solution:
a – b = 6, ab = 20
Cubing both sides,
(a – b)3 = (6)3
⇒  a3 – b3 – 3ab(a – b) = 216
⇒  a3 – b3 – 3 x 20 x 6 = 216
⇒  a3 – b3 – 360 = 216
⇒  a3 -b3 = 216 + 360 = 576
∴ a3 – b3 = 576

Question 6.
If x = -2 and y = 1, by using an identity find the value of the following:

Solution:

Hope given RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.4 are helpful to complete your math homework.

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## RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.1

Other Exercises

Question 1.
Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer:

Solution:
(i) 3x2 – 4x + 15,
(ii) y2 + 2$$\sqrt { 3 }$$ are polynomial is one variable. Others are not polynomial or polynomials in one variable.

Question 2.
Write the coefficient of x2 in each of the following:

Solution:
Coefficient of x2,
in (i) is 7
in (ii) is 0 as there is no term of x2 i.e. 0 x2

Question 3.
Write the degrees of each of the following polynomials:
(i) 7x3 + 4x2 – 3x + 12
(ii) 12 – x + 2x3
(iii) 5y – $$\sqrt { 2 }$$
(iv) 7
(v) 0
Solution:
(i) Degree of the polynomial 7x3 + 4x2 – 3x + 12 is 3
(ii) Degree of the polynomial 12 – x + 2x3 is 3
(iii) Degree of the polynomial 5y – $$\sqrt { 2 }$$is 1
(iv) Degree of the polynomial 7 is 0
(v) Degree of the polynomial 0 is 0 undefined.

Question 4.
(i) x + x2 + 4
(ii) 3x – 2
(iii) 2x + x2 [NCERT]
(iv) 3y
(v) t2 + 1
(v) 7t4 + 4t3 + 3t – 2
Solution:
(i)  x + x2 + 4 It is a quadratic polynomial.
(ii) 3x – 2 : It is a linear polynomial.
(iii) 2x + x2: It is a quadratic polynomial.
(iv) 3y It is a linear polynomial.
(v) t2+ 1 It is a quadratic polynomial.
(vi) 7t4 + 4t3 + 3t – 2 It is a biquadratic polynomial.

Question 5.
Classify the following polynomials as polynomials in one-variable, two-variables etc.
(i) x2-xy +7y2
(ii) x2 – 2tx + 7t2 – x + t
(iii) t3 -3t2 + 4t-5
(iv) xy + yz + zx
Solution:
(i) x2 – xy + 7y2: It is a polynomial in two j variables x, y.
(ii) x2 – 2tx + 7t2 – x + t: It is a polynomial in two variables in x, t.
(iii) t3 – 3t2 + 4t – 5 : It is a polynomial in one variable in t.
(iv) xy +yz + zx : It is a polynomial in 3 variables in x, y and

Question 6.
Identify polynomials in the following:

Solution:

Question 7.
Identify constant, linear, quadratic and cubic polynomials from the following polynomials:

Solution:
(i) f(x) = 0 : It is a constant polynomial as it has no variable.
(ii) g(x) = 2x3 – 7x + 4 : It is a cubic polynomial.
(iii) h(x) = -3x + $$\frac { 1 }{ 2 }$$ : It is a linear polynomial.
(iv) p(x) = 2x2 – x + 4 : It is a quadratic polynomial.
(v) q(x) = 4x + 3 : It is linear polynomial.
(vi) r(x) = 3x3 + 4x2 + 5x – 7 : It is a cubic polynomial.

Question 8.
Give one example each of a binomial of degree 35 and of a monomial of degree 100.   [NCERT]
Solution:
Example of a binomial of degree 35 = 9x35 + 16
Example of a monomial of degree 100 = 2y100

Hope given RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.1 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

## RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.5

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.5

Question 1.
Find the following products:
(i) (3x + 2y + 2z) (9x2 + 4y2 + 4z2 – 6xy – 4yz – 6zx)
(ii) (4x -3y + 2z) (16x2 + 9y2+ 4z2 + 12xy + 6yz – 8zx)
(iii) (2a – 3b – 2c) (4a2 + 9b2 + 4c2 + 6ab – 6bc + 4ca)
(iv) (3x -4y + 5z) (9x2 + 16y2 + 25z2 + 12xy- 15zx + 20yz)
Solution:
(i) (3x + 2y + 2z) (9x2 + 4y2 + 4z2 – 6xy – 4yz – 6zx)
= (3x + 2y + 2z) [(3x)2 + (2y)2 + (2z)2 – 3x x 2y + 2y x 2z + 2z x 3x]
= (3x)3 + (2y)3  + (2z)3 – 3 x 3x x 2y x 2z
= 27x3 + 8y3 + 8Z3 – 36xyz
(ii) (4x – 3y + 2z) (16x2 + 9y2 + 4z2 + 12xy + 6yz – 8zx)
= (4x -3y + 2z) [(4x)2 + (-3y)2 + (2z)2 – 4x x (-3y + (3y) x (2z) – (2z x 4x)]
= (4x)3 + (-3y)3 + (2z)3 – 3 x 4x x (-3y) x (2z)
= 64x3 – 27y3 + 8z3 + 72xyz
(iii) (2a -3b- 2c) (4a2 + 9b2 + 4c2 + 6ab – 6bc + 4ca)
= (2a -3b- 2c) [(2a)2 + (3b)2 + (2c)2 – 2a x (-3b) – (-3b) x (-2c) – (-2c) x 2a]
= (2a)3 + (3b)3 + (-2c)3 -3x 2a x (-3 b) (-2c)
= 8a3 – 21b3 -8c3 – 36abc
(iv) (3x – 4y + 5z) (9x2 + 16y2 + 25z2 + 12xy – 15zx + 20yz)
= [3x + (-4y) + 5z] [(3x)2 + (-4y)2 + (5z)2 – 3x x (-4y) -(-4y) (5z) – 5z x 3x]
= (3x)3 + (-4y)3 + (5z)3 – 3 x 3x x (-4y) (5z)
= 27x3 – 64y3 + 125z3 + 180xyz

Question 2.
Evaluate:

Solution:

Question 3.
If x + y + z = 8 and xy + yz+ zx = 20, find the value of x3 + y3 + z3 – 3xyz.
Solution:
We know that
x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 -xy -yz – zx)
Now, x + y + z = 8
Squaring, we get
(x + y + z)2 = (8)2
x2 + y2 + z2 + 2(xy + yz + zx) = 64
⇒ x2 + y2 + z2 + 2 x 20 = 64
⇒  x2 + y2 + z2 + 40 = 64
⇒  x2 + y2 + z2 = 64 – 40 = 24
Now,
x3 + y3 + z3 – 3xyz = (x + y + z) [x2 + y2 + z2 – (xy + yz + zx)]
= 8(24 – 20) = 8 x 4 = 32

Question 4.
If a +b + c = 9 and ab + bc + ca = 26, find the value of a3 + b3 + c3 – 3abc.
Solution:
a + b + c = 9, ab + be + ca = 26
Squaring, we get
(a + b + c)2 = (9)2
a2 + b2 + c2 + 2 (ab + be + ca) = 81
⇒  a2 + b2 + c2 + 2 x 26 = 81
⇒ a2 + b2 + c2 + 52 = 81
∴  a2 + b2 + c2 = 81 – 52 = 29
Now, a3 + b3 + c3 – 3abc = (a + b + c) [(a2 + b2 + c2 – (ab + bc + ca)]
= 9[29 – 26]
= 9 x 3 = 27

Question 5.
If a + b + c = 9, and a2 + b2 + c2 = 35, find the value of a3 + b3 + c3 – 3abc.
Solution:
a + b + c = 9
Squaring, we get
(a + b + c)2 = (9)2
⇒  a2 + b2 + c2 + 2 (ab + be + ca) = 81
⇒  35 + 2(ab + bc + ca) = 81
2(ab + bc + ca) = 81 – 35 = 46
∴  ab + bc + ca = $$\frac { 46 }{ 2 }$$ = 23
Now, a3 + b3 + c3 – 3abc
= (a + b + c) [a2 + b2 + c2 – (ab + bc + ca)]
= 9[35 – 23] = 9 x 12 = 108

Hope given RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.5 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

## RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.1

Question 1.
Evaluate each of the following using identities:
(i) (2x –$$\frac { 1 }{ x }$$)2
(ii)  (2x + y) (2x – y)
(iii) (a2b – b2a)2
(iv) (a – 0.1) (a + 0.1)
(v) (1.5.x2 – 0.3y2) (1.5x2 + 0.3y2)
Solution:

Question 2.
Evaluate each of the following using identities:
(i) (399)2
(ii) (0.98)2
(iii) 991 x 1009
(iv) 117 x 83
Solution:

Question 3.
Simplify each of the following:

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.
If 9x2 + 25y2 = 181 and xy = -6, find the value of 3x + 5y.
Solution:
9x2 + 25y2 = 181, and xy = -6
(3x + 5y)2 = (3x)2 + (5y)2 + 2 x 3x + 5y
⇒ 9X2 + 25y2 + 30xy
= 181 + 30 x (-6)
= 181 – 180 = 1
= (±1 )2
∴ 3x + 5y = ±1

Question 8.
If 2x + 3y = 8 and xy = 2, find the value of 4X2 + 9y2.
Solution:
2x + 3y = 8 and xy = 2
Now, (2x + 3y)2 = (2x)2 + (3y)2 + 2 x 2x x 3y
⇒  (8)2 = 4x2 + 9y2 + 12xy
⇒ 64 = 4X2 + 9y2 + 12 x 2
⇒ 64 = 4x2 + 9y2 + 24
⇒ 4x2 + 9y2 = 64 – 24 = 40
∴ 4x2 + 9y2 = 40

Question 9.
If 3x -7y = 10 and xy = -1, find the value of 9x2 + 49y2
Solution:
3x – 7y = 10, xy = -1
3x -7y= 10
Squaring both sides,
(3x – 7y)2 = (10)2
⇒ (3x)2 + (7y)2 – 2 x 3x x 7y = 100
⇒  9X2 + 49y2 – 42xy = 100
⇒  9x2 + 49y2 – 42(-l) = 100
⇒ 9x2 + 49y2 + 42 = 100
∴ 9x2 + 49y2 = 100 – 42 = 58

Question 10.
Simplify each of the following products:

Solution:

Question 11.

Solution:

Question 12.

Solution:

Question 13.
Simplify each of the following products:

Solution:

Question 14.
Prove that a2 + b2 + c2 – ab – bc – ca is always non-negative for all values of a, b and c.
Solution:

∵  The given expression is sum of these squares
∴ Its value is always positive Hence the given expression is always non-­negative for all values of a, b and c

Hope given RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.1 are helpful to complete your math homework.

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## RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.3

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.3

Other Exercises

Factorize:
Question 1.
64a3 + 125b3 + 240a2b + 300ab2
Solution:
64a3 + 125b3 + 240a2b + 300ab2
= (4a)3 + (5b)3 + 3 x (4a)2 x 5b + 3(4a) + (5b)2
= (4a + 5b)3
= (4a + 5b) (4a + 5b) (4a + 5b)

Question 2.
125x3 – 27y3 – 225x2y + 135xy2
Solution:
125x3 – 27y3 – 225x2y + 135xy2
= (5x)3 – (3y)3 – 3 x (5x)2 x (3y) + 3- x 5x x (3y)2
= (5x – 3y)3
=
(5x – 3y) (5x – 3y) (5x – 3y)

Question 3.

Solution:

Question 4.
8x3 + 27y3 + 36x2y + 54xy2
Solution:
8x3 + 27y3 + 16x2y + 54xy2
= (2x)3 + (3y)3 + 3 x (2x)2 x 3y  +  3 x 2x x (3y)2
= (2x + 3y)3
= (2x + 3y) (2x + 3y) (2x + 3y)

Question 5.
a3 – 3a2b + 3ab2 – b3 + 8
Solution:
a3 – 3a2b + 3ab2 – b3 + 8
= (a – b)3 + (2)3
= (a – b + 2) [(a -b)2– (a – b) x 2 + (2)2]
= (a- b + 2) (a2 + b2 -2ab – 2a + 2b + 4)

Question 6.
x3 + 8y3 + 6x2y + 12xy2
Solution:
x3 + 8y3 + 6x2y + 12xy2
= (x)3 + (2y)3 + 3 x x2x 2y + 3 x x x (2y)2
= (x + 2y)3
= (x + 2y) (x + 2y) (x + 2y)

Question 7.
8x3 + y3 + 12x2y + 6xy2
Solution:
8x3 + y3 + 12x2y + 6xy2
= (2x)3 + (y)3 + 3 x (2x)2 x y + 3 x 2x x y2
= (2x + y)3
= (2x + y) (2x + y) (2x + y)

Question 8.
8a3 + 27b3 + 36a2b + 54ab2
Solution:
8a3 + 27b3 + 16a2b + 54ab2
= (2a)3 + (3b)3 + 3 x (2a)x 3b + 3 x 2a x (3b)2
= (2a + 3b)3
= (2a + 3b) (2a + 3b) (2a + 3b)

Question 9.
8a3 – 27b3 – 36a2b + 54ab2
Solution:
8a3 – 27b3 – 36a2b + 54ab2
= (2a)3 – (3b)3 – 3 x (2a)2 x 3b + 3 x 2a x (3b)2
= (2a – 3b))3
= (2a – 3b) (2a – 3b) (2a – 3b)

Question 10.
x3 – 12x(x – 4) – 64
Solution:
x3 – 12x(x – 4) – 64
= x3 – 12x2 + 48x – 64
= (x)3 – 3 x x2 x 4 + 3 x x x (4)2– (4)3
= (x – 4)3
= (x – 4) (x – 4) (x – 4)

Question 11.
a3x3 – 3a2bx2 + 3ab2x – b3
Solution:
a3x3 – 3a2bx2 + 3ab2x – b3
= (ax)3 – 3 x (ax)2 x  b + 3 x ax x (b)2– (b)3
= (ax – b)3
= (ax – b) (ax – b) (ax – b)

Hope given RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.3 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

## RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities MCQS

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.
If a + b + c = 9 and  ab + bc + ca = 23, then a2 + b2 + c2 =
(a) 35
(b) 58
(c) 127
(d) none of these
Solution:
a + b + c = 9, ab + bc + ca = 23
Squaring,
(a + b+ c) = (9)2
a2 + b2 + c2 + 2 (ab + bc + ca) = 81
⇒ a2 + b2 + c2 + 2 x 23 = 81
⇒  a2 + b2+ c2 + 46 = 81
⇒  a2 + b2+ c2 = 81 – 46 = 35   (a)

Question 9.
(a – b)3 + (b – c)3 + (c – a)3 =
(a) (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
(d) (a -b)(b- c) (c – a)
(c) 3(a – b) (b – c) (c – a)
(d) none of these
Solution:
(a – b)3 + (b- c)3 + (c- a)3
∵ a – b + b – c + c – a = 0
∴ (ab)3 + (b – c)3 + (c – a)3
= 3
(a -b)(b- c) (c – a)        (c)

Question 10.

Solution:

Question 11.
If a – b = -8 and ab = -12 then a3 – b3 =
(a) -244
(b) -240
(c) -224
(d) -260
Solution:
a – b = -8, ab = -12
(a – b)3 = a3 – b3 – 3ab (a – b)
(-8)3 = a3 – b3 – 3 x (-12) (-8)
-512 = a3-b3– 288
a3 – b3 = -512 + 288 = -224      (c)

Question 12.
If the volume of a cuboid is 3x2 – 27, then its possible dimensions are
(a) 3, x2, -27x
(b) 3, x – 3, x + 3
(c) 3, x2, 27x
(d) 3, 3, 3
Solution:
Volume = 3x2 -27 = 3(x2 – 9)
= 3(x + 3) (x – 3)
∴ Dimensions are   = 3, x – 3, x   +  3        (b)

Question 13.
75 x 75 + 2 x 75 x 25    + 25 x 25 is equal to
(a) 10000
(b) 6250
(c) 7500
(d) 3750
Solution:

Question 14.
(x – y) (x + y)(x2 + y2) (x4 + y4) is equal to
(a) x16 – y16
(b) x8 – y8
(c) x8 + y8
(d) x16 + y16
Solution:

Question 15.

Solution:

Question 16.

Solution:

Question 17.

Solution:

Question 18.

Solution:

Question 19.
If a2 + b2 + c2 – ab – bc – ca = 0, then
(a) a + b = c
(b) b + c = a
(c) c + a = b
(d) a = b = c
Solution:
a2 + b2 + c2 – ab – bc – ca = 0
2 {a2 + b2 + c2 – ab – be – ca) = 0 (Multiplying by 2)
⇒  2a2 + 2b2 + 2c2– 2ab – 2bc – 2ca = 0
⇒  a2 + b2 – 2ab + b2 + c2 – 2bc + c2 + a2 – 2ca = 0
⇒  (a – b)2 + (b – c)2 + (c – a)2 = 0
(a – b)2 = 0, then a – b = 0
⇒ a = b
Similarly, (b – c)2 = 0, then
b-c = 0
⇒ b = c
and (c – a)2 = 0, then c-a = 0
⇒ c = a
∴ a = b – c           (d)

Question 20.

Solution:

Question 21.

Solution:

Question 22.
If a + b + c = 9 and ab + bc + ca = 23, then a3 + b3 + c3 – 3 abc =
(a) 108
(b) 207
(c) 669
(d) 729
Solution:
a3 + b3 + c3 – 3abc
= (a + b + c) [a2 + b2 + c2 – (ab + bc + ca)
Now, a + b + c = 9
Squaring,
a2 + b2 + c2 + 2 (ab + be + ca) = 81
⇒  a2 + b2 + c2 + 2 x 23 = 81
⇒  a2 + b2 + c2 + 46 = 81
⇒  a2 + b2 + c2 = 81 – 46 = 35
Now, a3 + b3 + c3 – 3 abc = (a + b + c) [(a2 + b2 + c2) – (ab + bc + ca)
= 9[35 -23] = 9 x 12= 108                     (a)

Question 23.

Solution:

Question 24.
The product (a + b) (a – b) (a2 – ab + b2) (a2 + ab + b2) is equal to
(a) a6 +   b6
(b) a6 – b6
(c) a3 – b3
(d) a3 + b3
Solution:
(a + b) (a – b) (a2 – ab + b2) (a2 + ab +b2)
= (a + b) (a2-ab + b2) (a-b) (a2 + ab + b2)
= (a3 + b3) (a3 – b3)
= (a3)2 – (b3)2 = a6 – b6   (b)

Question 25.
The product (x2 – 1) (x4 + x2 + 1) is equal to
(a) x8 –   1
(b) x8 + 1
(c) x6 –   1
(d) x6 + 1
Solution:
(x2 – 1) (x4 + x2 + 1)
= (x2)3 – (1)3 = x6 – 1                            (c)

Question 26.

Solution:

Question 27.

Solution:

Hope given RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

## RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions MCQS

Other Exercises

Mark the correct alternative in each of the following:
Question 1.
The factors of x3 – x2y -xy2 + y3 are
(a) (x + y) (x2 -xy + y2)
(b) (x+y)(x2 + xy + y2)
(c) (x + y)2 (x – y)
(d) (x – y)2 (x + y)
Solution:
x3 – x2y – xy2 + y3
= x3 + y3 – x2y – xy2
= (x + y) (x2 -xy + y2)- xy(x + y)
= (x + y) (x2 – xy + y2 – xy)
= (x + y) (x2 – 2xy + y2)
= (x + y) (x – y)2         (d)

Question 2.
The factors of x3 – 1 +y3 + 3xy are
(a) (x – 1 + y)  (x2 + 1 + y2 + x + y – xy)
(b) (x + y + 1)  (x2 + y2 + 1- xy – x – y)
(c) (x – 1 + y)   (x2 – 1 – y+ x + y + xy)
(d) 3(x + y – 1) (x2 + y2 – 1)
Solution:
x3 – 1 + y3 + 3xy
= (x)3 + (-1)3 + (y)3 – 3 x  x  x (-1) x y
= (x – 1 + y) (x2 + 1 + y2 + x + y – xy)
= (x- 1 + y) (x2+ 1 + y2 + x + y – xy)      (a)

Question 3.
The factors of 8a3 + b3 – 6ab + 1 are
(a) (2a + b – 1) (4a2 + b2 + 1 – 3ab – 2a)
(b) (2a – b + 1) (4a2 + b2 – 4ab + 1 – 2a + b)
(c) (2a + b+1) (4a2 + b2 + 1 – 2ab – b – 2a)
(d) (2a – 1 + b)(4a2 + 1 – 4a – b – 2ab)
Solution:
8a3 + b3 – 6ab + 1
= (2a)3 + (b)3 + (1)3 – 3 x 2a x b x 1
= (2a + b + 1) [(2a)2 + b2+1-2a x b- b x 1 – 1 x 2a]
= (2a + b + 1) (4a2 + b2+1-2ab-b- 2a)            (c)

Question 4.
(x + y)3 – (x – v)3 can be factorized as
(a) 2y (3x2 + y2)
(b) 2x (3x2 + y2)
(c) 2y (3y2 + x2)
(d) 2x (x2 + 3y2)
Solution:
(x + y)3 – (x – y)3
= (x + y -x + y) [(x + y)2 + (x +y) (x -y) + (x – y)2]
= 2y(x2 + y2 + 2xy + x2-y2 + x2+y2 – 2xy)
= 2y(3x2 + y2)          (a)

Question 5.
The expression (a – b)3 + (b – c)3 + (c – a)3 can be factorized as
(a) (a -b) (b- c) (c – a)
(b) 3(a – b) (b – c) (c – a)
(c) -3(a – b) (b – c) (a – a)
(d) (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
Solution:
(a – b)3 + (b – c)3 + (c – a)3
Let a – b = x, b – a = y, c – a = z
∴ x3 + y3 + z3
x+y + z = a- b + b- c + c – a = 0
∴ x3 +y3 + z3 = 3xyz
(a – b)3 + (b – a)3 + (c – a)3
= 3 (a – b) (b – c) (c – a)        (b)

Question 6.

Solution:

Question 7.

Solution:

Question 8.
The factors of a2 – 1 – 2x – x2 are
(a) (a – x + 1) (a – x – 1)
(b) (a + x – 1) (a – x + 1)
(c) (a + x + 1) (a – x – 1)
(d) none of these
Solution:
a2 – 1- 2x – x2
⇒ a2 – (1 + 2x + x2)
= (a)2 – (1 + x)2
= (a + 1 + x) (a – 1 – x)                         (c)

Question 9.
The factors of x4 + x2 + 25 are
(a) (x2 + 3x + 5) (x2 – 3x + 5)
(b) (x2 + 3x + 5) (x2 + 3x – 5)
(c) (x2 + x + 5) (x2 – x + 5)
(d) none of these
Solution:
x4 + x2 + 25 = x4 + 25 +x2
= (x2)2 + (5)2 + 2 x x2 x 5- 9x2
= (x2 + 5)2 – (3x)2
= (x2 + 5 + 3x) (x2 + 5 – 3x)
= (x2 + 3x + 5) (x2 – 3x + 5)                 (a)

Question 10.
The factors of x2 + 4y2 + 4y – 4xy – 2x – 8 are
(a) (x – 2y – 4) (x – 2y + 2)
(b)  (x – y  +   2) (x – 4y – 4)
(c) (x + 2y – 4) (x + 2y + 2)
(d)    none of these
Solution:
x2 + 4y2 + 4y – 4xy – 2x – 8
⇒  x2 + 4y + 4y – 4xy – 2x – 8
= (x)2 + (2y)2– 2 x x x 2y + 4y-2x-8
= (x – 2y)2 – (2x – 4y) – 8
= (x – 2y)2 – 2 (x – 2y) – 8
Let x – 2y = a, then
a2– 2a – 8 = a2– 4a + 2a – 8
= a(a – 4) + 2(a – 4)
= (a-4) (a + 2)
= (x2 -2y-4) (x2 -2y + 2)                       (a)

Question 11.
The factors of x3 – 7x + 6 are
(a) x(x – 6) (x – 1)
(b) (x2 – 6) (x – 1)
(c) (x + 1) (x + 2) (x – 3)
(d) (x – 1) (x + 3) (x – 2)
Solution:
x-7x + 6= x3-1-7x + 7
= (x – 1) (x2 + x + 1) – 7(x – 1)
= (x – 1) (x2 + x + 1 – 7)
= (x – 1) (x2 + x – 6)
= (x – 1) [x2 + 3x – 2x – 6]
= (x – 1) [x(x + 3) – 2(x + 3)]
= (x – 1) (x+ 3) (x – 2)                           (d)

Question 12.
The expression x4 + 4 can be factorized as
(a) (x2 + 2x + 2) (x2 – 2x + 2)
(b) (x2 + 2x + 2) (x2 + 2x – 2)
(c) (x2 – 2x – 2) (x2 – 2x + 2)
(d) (x2 + 2) (x2 – 2)
Solution:
x4 + 4 = x4 + 4 + 4x2 – 4x2                (Adding and subtracting 4x2)
= (x2)2 + (2)2 + 2 x x2 x 2 – (2x)2
= (x2 + 2)2 – (2x)2
= (x2 + 2 + 2x) (x2 + 2 – 2x)                {∵ a2 – b2 = (a + b) (a – b)}
= (x2 + 2x + 2) (x2 – 2x + 2)                  (a)

Question 13.
If 3x = a + b + c, then the value of (x – a)3 + (x –    bf + (x – cf – 3(x – a) (x – b) (x – c) is
(a) a + b + c
(b) (a – b) {b – c) (c – a)
(c) 0
(d) none of these
Solution:
3x = a + b + c                                                                      .
⇒ 3x-a-b-c = 0
Now, (x – a)3+ (x – b)3 + (x – c)– 3(x – a) (x -b)  (x – c)
= {(x – a) + (x – b) + (x – c)} {(x – a)2 + (x – b)+ (x – c)2  – (x – a) (x – b) (x – b) (x – c) – (x – c) (x – a)}
= (x – a + x – b + x – c) {(x – a)2 + (x – b)2  + (x – c)2 – (x – a) (x – b) – (x – b) (x – c) – (x – c) (x – a)}
= (3x – a – b -c) {(x – a)2 + (x -b)2+ (x – c)2 – (x – a) (x – b) – (x – b) (x – c) – (x – c) (x – a)}
But 3x-a-b-c = 0, then
= 0 x {(x – a)2 + (x – b)2 + (x – c)2 – (x – a) (x – b) – (x – b) (x – c) – (x – c) (x – a)}
= 0                                                         (c)

Question 14.
If (x + y)3 – (x – y)3 – 6y(x2 – y2) = ky2, then k =
(a) 1
(b) 2
(c) 4
(d) 8
Solution:
(x + y)3 – (x – y)3 – 6y(x2 – y2) = ky2
LHS = (x + y)3 – (x – y)3 – 3 x (x + y) (x – y) [x + y – x + y]
= (x+y-x + y)3       {∵ a3 – b3 – 3ab (a – b) = a3 – b3}
= (2y)3 = 8y3
Comparing with ky3, k = 8                     (d)

Question 15.
If x3 – 3x2 + 3x – 7 = (x + 1) (ax2 + bx + c), then a + b + c =
(a) 4
(b) 12
(c) -10
(d) 3
Solution:
x3 – 3x2 + 3x + 7 = (x + 1) (ax2 + bx + c)
= ax3 + bx2 + cx + ax2 + bx + c
x3 – 3x2 + 3x – 7 = ax3 + (b + a)2 + (c + b)x + c
Comparing the coefficient,
a = 1
b + a = -3 ⇒ b+1=-3 ⇒ b = -3-1=-4
c + b = 3 ⇒ c- 4 = 3 ⇒ c = 3 + 4 = 7
a + b + c = 1- 4 + 7 = 8- 4 = 4             (a)

Hope given RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.