RD Sharma Solutions Class 9 Maths Chapter 4 – Algebraic Identities (Updated for 2021-22)

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RD Sharma Solutions Class 9 Maths Chapter 4

RD Sharma Solutions Class 9 Maths Chapter 4 Algebraic Identities: Exercise wise Solutions

RD Sharma Solutions Class 9 Chapter 4 Exercise 4.1

RD Sharma Solutions Class 9 Chapter 4 Exercise 4.2

RD Sharma Solutions Class 9 Chapter 4 Exercise 4.3

RD Sharma Solutions Class 9 Chapter 4 Exercise 4.4

Access answers of RD Sharma Solutions Class 9 Maths Chapter 4 – Algebraic Identities

RD Sharma Solutions Class 9 Chapter 4 Algebraic Identities Ex 4.1

Question 1.
Evaluate each of the following using identities:
(i) (2x –1x)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

Class 9 RD Sharma 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 =-162 = -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:

Class 9 RD Sharma Solutions Chapter 4 Algebraic Identities Ex 4.3

Question 1.
Find the cube of each of the following binomial expressions:

Solution:

Question 2.
If a + b = 10 and ab = 21, find the value of a³ + b³.
Solution:
a + b = 10, ab = 21
Cubing both sides,
(a + b)³ = (10)³
⇒ a³ + 6³ + 3ab (a + b) = 1000
⇒  a³ + b³ + 3 x 21 x 10 = 1000
⇒  a³ + b³ + 630 = 1000
⇒  a³ + b³ = 1000 – 630 = 370
∴ a³ + b³ = 370

Question 3.
If a – b = 4 and ab = 21, find the value of a³-b³.
Solution:
a – b = 4, ab= 21
Cubing both sides,
⇒ (a – A)³ = (4)³
⇒ a³ – b³ – 3ab (a – b) = 64
⇒ a³-i³-3×21 x4 = 64
⇒  a³ – 6³ – 252 = 64
⇒  a³ – 6³ = 64 + 252 =316
∴ a³ – b³ = 316

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.
If 2x + 3y = 13 and xy = 6, find the value of 8x³ + 21y³.
Solution:
2x + 3y = 13, xy = 6
Cubing both sides,
(2x + 3y)³ = (13)³
⇒ (2x)³ + (3y)³ + 3 x 2x x 3X2x + 3y) = 2197
⇒ 8x³ + 27y³ + 18xy(2x + 3y) = 2197
⇒ 8x³ + 27y³ + 18 x 6 x 13 = 2197
⇒ 8X³ + 27y³ + 1404 = 2197
⇒  8x³ + 27y³ = 2197 – 1404 = 793
∴ 8x³ + 27y³ = 793

Question 10.
If 3x – 2y= 11 and xy = 12, find the value of 27x³ – 8y³.
Solution:
3x – 2y = 11 and xy = 12 Cubing both sides,
(3x – 2y)³ = (11)³
⇒  (3x)³ – (2y)³ – 3 x 3x x 2y(3x – 2y) =1331
⇒  27x³ – 8y³ – 18xy(3x -2y) =1331
⇒   27x³ – 8y³ – 18 x 12 x 11 = 1331
⇒  27x³ – 8y³ – 2376 = 1331
⇒  27X³ – 8y³ = 1331 + 2376 = 3707
∴ 2x³ – 8y³ = 3707

Question 11.
Evaluate each of the following:
(i)  (103)³
(ii) (98)³
(iii) (9.9)³
(iv) (10.4)³
(v) (598)³
(vi) (99)³
Solution:
We know that (a + bf = a³ + b³ + 3ab(a + b) and (a – b)³= a³ – b³ – 3 ab(a – b)
Therefore,
(i)  (103)³ = (100 + 3)³
= (100)³ + (3)³ + 3 x 100 x 3(100 + 3)    {∵ (a + b)³ = a³ + b³ + 3ab(a + b)}
= 1000000 + 27 + 900 x 103
= 1000000 + 27 + 92700
= 1092727
(ii) (98)³ = (100 – 2)³
= (100)³ – (2)³ – 3 x 100 x 2(100 – 2)
= 1000000 – 8 – 600 x 98
= 1000000 – 8 – 58800
= 1000000-58808
= 941192
(iii) (9.9)³ = (10 – 0.1)³
= (10)³ – (0.1)³ – 3 X 10 X 0.1(10 – 0.1)
= 1000 – 0.001 – 3 x 9.9
= 1000 – 0.001 – 29.7
= 1000 – 29.701
= 970.299
(iv) (10.4)³ = (10 + 0.4)³
= (10)³ + (0.4)³ + 3 x 10 x 0.4(10 + 0.4)
= 1000 + 0.064 + 12(10.4)
= 1000 + 0.064 + 124.8 = 1124.864
(v) (598)³ = (600 – 2)³
= (600)³ – (2)³ – 3 x 600 x 2 x (600 – 2)
= 216000000 – 8 – 3600 x 598
= 216000000 – 8 – 2152800
= 216000000 – 2152808
= 213847192
(vi) (99)³ = (100 – 1)³
= (100)³ – (1)³ – 3 x 100 x 1 x (100 – 1)
= 1000000 – 1 – 300 x 99
= 1000000 – 1 – 29700
= 1000000 – 29701
= 970299

Question 12.
Evaluate each of the following:
(i)  111³ – 89³
(ii) 46³ + 34³
(iii) 104³ + 96³
(iv) 93³ – 107³
Solution:
We know that a³ + b³ = (a + bf – 3ab(a + b) and a³ – b³ = (a – bf + 3 ab(a – b)
(i) 111³ – 89³
= (111 – 89)³ + 3 x ill x 89(111 – 89)
= (22)³ + 3 x 111 x 89 x 22
= 10648 + 652014 = 662662
(OR)
(a + b)³ – (a – b)³ = 2(b³ + 3a²b)
= 111³ – 89³ = (100 + 11)³ – (100 – 11)³
= 2(11³ + 3 x 100² x 11]
= 2(1331 + 330000]
= 331331 x 2 = 662662
(a + b)³ + (a- b)³ = 2(b³ + 3ab²)
(ii) 46³ + 34³ = (40 + 6)³ + (40 – 6)³
= 2[(40)³ + 3 x 40 x 6²]
= 2[64000 + 3 x 40 x 36]
= 2[64000 + 4320]
= 2 x 68320 = 136640
(iii) 104³ + 96³ = (100 + 4)³ + (100 – 96)³
= 2 [a³ + 3 ab²]
= 2[(100)³ + 3 x 100 x (4)²]
= 2[ 1000000 + 300 x 16]
= 2[ 1000000 + 4800]
= 1004800 x 2 = 2009600
(iv) 93³ – 107³ = -[(107)³ – (93)³]
= -[(100 + If – (100 – 7)³]
= -2[b³ + 3a²b)]
= -2[(7)³ + 3(100)² x 7]
= -2(343 + 3 x 10000 x 7]
= -2[343 + 210000]
= -2[210343] = -420686

Question 13.

Solution:

Question 14.
Find the value of 27X³ + 8y³ if
(i) 3x + 2y = 14 and xy = 8
(ii) 3x + 2y = 20 and xy = 149
Solution:

Question 15.
Find the value of 64x³ – 125z³, if 4x – 5z = 16 and xz = 12.
Solution:
4x – 5z = 16, xz = 12
Cubing both sides,
(4x – 5z)³ = (16)³
⇒ (4x)³ – (5y)³ – 3 x 4x x 5z(4x – 5z) = 4096
⇒ 64x³ – 125z³ – 3 x 4 x 5 x xz(4x – 5z) = 4096
⇒  64x³ – 125z³ – 60 x 12 x 16 = 4096
⇒ 64x³ – 125z³ – 11520 = 4096
⇒  64x³ – 125z³ = 4096 + 11520 = 15616

Question 16.

Solution:

Question 17.
Simplify each of the following:

Solution:

Question 18.


Solution:

Question 19.

Solution:

RD Sharma Mathematics 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:

RD Sharma Solutions Class 9 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 = 462 = 23
Now, a³ + b³ + c³ – 3abc
= (a + b + c) [a² + b² + c² – (ab + bc + ca)]
= 9[35 – 23] = 9 x 12 = 108

Class 9 Maths Chapter 4 Algebraic Identities RD Sharma Solutions 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:

Algebraic Identities Class 9 RD Sharma Solutions 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
∴ (a – b)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:

Important Topics: RD Sharma Solutions Class 9 Maths Chapter 4 

Important topics always let you run a thorough check on the titles that are there in the chapters. So, the important topics which the chapter includes are-

• Algebraic Identities Introduction
• Identity for the square of a trinomial
• Sum and difference of cubes Identity

Here is everything which students need to smoothly finish their CBSE Class 9 Mathematics syllabus. Start with CBSE Class 9 to build a solid future ahead. For any doubts, ask in the comments. 

FAQs on RD Sharma Solutions Class 9 Maths Chapter 4 – Algebraic Identities