## RD Sharma Class 9 Solutions 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 –\(\frac { 1 }{ x }\))^{2}

(ii)Â (2x + y) (2x – y)

(iii) (a^{2}b – b^{2}a)^{2}

(iv) (a – 0.1) (a + 0.1)

(v) (1.5.x^{2} – 0.3y^{2}) (1.5x^{2} + 0.3y^{2})

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 9x^{2} + 25y^{2} = 181 and xy = -6, find the value of 3x + 5y.

Solution:

9x^{2} + 25y^{2} = 181, and xy = -6

(3x + 5y)^{2} = (3x)^{2} + (5y)^{2} + 2 x 3x + 5y

â‡’ 9X^{2} + 25y^{2} + 30xy

= 181 + 30 x (-6)

= 181 – 180 = 1

= (Â±1 )^{2}

âˆ´ 3x + 5y = Â±1

Question 8.

If 2x + 3y = 8 and xy = 2, find the value of 4X^{2} + 9y^{2}.

Solution:

2x + 3y = 8 and xy = 2

Now, (2x + 3y)^{2} = (2x)^{2} + (3y)^{2} + 2 x 2x x 3y

â‡’Â (8)^{2} = 4x^{2} + 9y^{2} + 12xy

â‡’ 64 = 4X^{2} + 9y^{2} + 12 x 2

â‡’ 64 = 4x^{2} + 9y^{2} + 24

â‡’ 4x^{2} + 9y^{2} = 64 – 24 = 40

âˆ´ 4x^{2} + 9y^{2} = 40

Question 9.

If 3x -7y = 10 and xy = -1, find the value of 9x^{2} + 49y^{2}

Solution:

3x – 7y = 10, xy = -1

3x -7y= 10

Squaring both sides,

(3x – 7y)^{2} = (10)^{2
}â‡’ (3x)^{2} + (7y)^{2} – 2 x 3x x 7y = 100

â‡’Â 9X^{2} + 49y^{2} – 42xy = 100

â‡’Â 9x^{2} + 49y^{2} – 42(-l) = 100

â‡’ 9x^{2} + 49y^{2} + 42 = 100

âˆ´ 9x^{2} + 49y^{2} = 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 a^{2} + b^{2} + c^{2} – ab – bc – ca is always non-negative for all values of a, b and c.

Solution:

âˆµÂ The given expression is sum of these squares

âˆ´ Its value is always positive Hence the given expression is always non-Ânegative 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 a^{2} + b^{2} + c^{2} = 16, find the value of ab + be + ca.

Solution:

a + b+ c = 0

Squaring both sides,

(a + b + c)^{2} = 0

â‡’ a^{2} + b^{2} + c^{2} + 2(ab + bc + ca) = 0

16 + 2(ab + bc + c) =Â 0

â‡’ 2(ab + bc + ca) = -16

â‡’Â ab + bc + ca =-\(\frac { 16 }{ 2 }\) = -8

âˆ´ ab + bc + ca = -8

Question 3.

If a^{2} + b^{2} + c^{2} = 16 and ab + bc + ca = 10, find the value of a + b + c.

Solution:

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab + bc + ca)

= 16 + 2 x 10

= 16 + 20 = 36

= (Â±6)^{2
}âˆ´ a + b + c = Â±6

Question 4.

If a + b + c = 9 and ab + bc + ca = 23, find the value of a^{2} + b^{2} + c^{2}.

Solution:

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab + bc + ca)

â‡’ (9)^{2} = a^{2} + b^{2} + c^{2} + 2 x 23

â‡’Â 81= a^{2} + b^{2} + c^{2} + 46

â‡’Â Â a^{2} + b^{2} + c^{2} = 81 – 46 = 35

âˆ´Â a^{2} + b^{2} + c^{2} = 35

Question 5.

Find the value of 4x^{2} + y^{2} + 25z^{2} + 4xy – 10yz – 20zx when x = 4, y = 3 and z = 2.

Solution:

x = 4, y – 3, z = 2

â‡’ 4x^{2} + y^{2} + 25z^{2} + 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)Â (a + b + c)^{2} + (a – b + c)^{2}

(ii) (a + b + c)^{2} –Â (a – b + c)^{2
}(iii) (a + b + c)^{2} +Â Â (a – b + c)^{2} + (a + b – c)^{2
}(iv) (2x + p – c)^{2} – (2x – p + c)^{2}

(v) (x^{2} + y^{2} – z^{2})^{2} – (x^{2} – y^{2} + z^{2})^{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 a^{3} + b^{3}.

Solution:

a + b = 10, ab = 21

Cubing both sides,

(a + b)^{3} = (10)^{3
}â‡’ a^{3} + 6^{3} + 3ab (a + b) = 1000

â‡’Â a^{3} + b^{3} + 3 x 21 x 10 = 1000

â‡’Â a^{3} + b^{3} + 630 = 1000

â‡’Â a^{3} + b^{3} = 1000 – 630 = 370

âˆ´ a^{3} + b^{3} = 370

Question 3.

If a – b = 4 and ab = 21, find the value of a^{3}-b^{3}.

Solution:

a – b = 4, ab= 21

Cubing both sides,

â‡’ (a – A)^{3} = (4)^{3
}â‡’ a^{3} – b^{3} – 3ab (a – b) = 64

â‡’ a^{3}-i^{3}-3×21 x4 = 64

â‡’Â a^{3} – 6^{3} – 252 = 64

â‡’Â a^{3} – 6^{3} = 64 + 252 =316

âˆ´ a^{3} – b^{3} = 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 8x^{3} + 21y^{3}.^{
}Solution:

2x + 3y = 13, xy = 6

Cubing both sides,

(2x + 3y)^{3} = (13)^{3
}â‡’ (2x)^{3} + (3y)^{3} + 3 x 2x x 3X2x + 3y) = 2197

â‡’ 8x^{3} + 27y^{3} + 18xy(2x + 3y) = 2197

â‡’ 8x^{3} + 27y^{3} + 18 x 6 x 13 = 2197

â‡’ 8X^{3} + 27y^{3} + 1404 = 2197

â‡’Â 8x^{3} + 27y^{3} = 2197 – 1404 = 793

âˆ´ 8x^{3} + 27y^{3} = 793

Question 10.

If 3x – 2y= 11 and xy = 12, find the value of 27x^{3} – 8y^{3}.

Solution:

3x – 2y = 11 and xy = 12 Cubing both sides,

(3x – 2y)^{3} = (11)^{3
}â‡’Â (3x)^{3} – (2y)^{3} – 3 x 3x x 2y(3x – 2y) =1331

â‡’Â 27x^{3} – 8y^{3} – 18xy(3x -2y) =1331

â‡’Â Â 27x^{3} – 8y^{3} – 18 x 12 x 11 = 1331

â‡’Â 27x^{3} – 8y^{3} – 2376 = 1331

â‡’Â 27X^{3} – 8y^{3} = 1331 + 2376 = 3707

âˆ´ 2x^{3} – 8y^{3} = 3707

Question 11.

Evaluate each of the following:

(i)Â (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 = a^{3} + b^{3} + 3ab(a + b) and (a – b)^{3}= a^{3} – b^{3} – 3 ab(a – b)

Therefore,

(i)Â (103)^{3} = (100 + 3)^{3
}= (100)^{3} + (3)^{3} + 3 x 100 x 3(100 + 3)Â Â {âˆµ (a + b)^{3} = a^{3} + b^{3} + 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)Â 111^{3} – 89^{3}

(ii) 46^{3} + 34^{3
}(iii) 104^{3} + 96^{3}

(iv) 93^{3} – 107^{3
}Solution:

We know that a^{3} + b^{3} = (a + bf – 3ab(a + b) and a^{3} – b^{3} = (a – bf + 3 ab(a – b)

(i) 111^{3} – 89^{3
}= (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(b^{3} + 3a^{2}b)

= 111^{3} – 89^{3} = (100 + 11)^{3} – (100 – 11)^{3
}= 2(11^{3} + 3 x 100^{2} x 11]

= 2(1331 + 330000]

= 331331 x 2 = 662662

(a + b)^{3} + (a- b)^{3} = 2(b^{3} + 3ab^{2})

(ii) 46^{3} + 34^{3} = (40 + 6)^{3} + (40 – 6)^{3
}= 2[(40)^{3} + 3 x 40 x 6^{2}]

= 2[64000 + 3 x 40 x 36]

= 2[64000 + 4320]

= 2 x 68320 = 136640

(iii) 104^{3} + 96^{3} = (100 + 4)^{3} + (100 – 96)^{3
}= 2 [a^{3} + 3 ab^{2}]

= 2[(100)^{3} + 3 x 100 x (4)^{2}]

= 2[ 1000000 + 300 x 16]

= 2[ 1000000 + 4800]

= 1004800 x 2 = 2009600

(iv) 93^{3} – 107^{3} = -[(107)^{3} – (93)^{3}]

= -[(100 + If – (100 – 7)^{3}]

= -2[b^{3} + 3a^{2}b)]

= -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 27X^{3} + 8y^{3} if

(i) 3x + 2y = 14 and xy = 8

(ii) 3x + 2y = 20 and xy = \(\frac { 14 }{ 9 }\)

Solution:

Question 15.

Find the value of 64x^{3} – 125z^{3}, if 4x – 5z = 16 and xz = 12.

Solution:

4x – 5z = 16, xz = 12

Cubing both sides,

(4x – 5z)^{3} = (16)^{3
}â‡’ (4x)^{3} – (5y)^{3} – 3 x 4x x 5z(4x – 5z) = 4096

â‡’ 64x^{3} – 125z^{3} – 3 x 4 x 5 x xz(4x – 5z) = 4096

â‡’Â 64x^{3} – 125z^{3} – 60 x 12 x 16 = 4096

â‡’ 64x^{3} – 125z^{3} – 11520 = 4096

â‡’Â 64x^{3} – 125z^{3} = 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) (9X^{2} – 6xy + Ay^{2})

(ii) (4x – 5y) (16x^{2} + 20xy + 25y^{2})

(iii) (7p^{4} + q) (49p^{8} – 7p^{4}q + q^{2})

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 a^{2} – ab + b^{2} and a^{2} + ab + b^{2}.

Solution:

a + b = 10, ab = 16 Squaring,

(a + b)^{2} = (10)^{2
}â‡’ a^{2} + b^{2} + lab = 100

â‡’ a^{2} + b^{2} + 2 x 16 = 100

â‡’Â a^{2} + b^{2} + 32 = 100

âˆ´ a^{2} + b^{2} = 100 – 32 = 68

Now, a^{2} – ab + b^{2} = a^{2} + b^{2} – ab = 68 – 16 = 52

and a^{2} + ab + b^{2} = a^{2} + b^{2} + ab = 68 + 16 = 84

Question 4.

If a + b = 8 and ab = 6, find the value of a^{3} + b^{3}.

Solution:

a + b = 8, ab = 6

Cubing both sides,

(a + b)^{3} = (8)3

â‡’ a^{3} + b^{3} + 3 ab{a + b) = 512

â‡’Â a^{3} + b^{3} + 3 x 6 x 8 = 512_{
}â‡’Â a^{3} + b^{3} + 144 = 512

â‡’Â a^{3} + b^{3} = 512 – 144 = 368

âˆ´ a^{3} + b^{3} = 368

Question 5.

If a – b = 6 and ab = 20, find the value of a^{3}-b^{3}.

Solution:

a – b = 6, ab = 20

Cubing both sides,

(a – b)3 = (6)^{3}

â‡’Â a^{3}Â – b^{3} – 3ab(a – b) = 216

â‡’Â a^{3} – b^{3} – 3 x 20 x 6 = 216

â‡’Â a^{3} – b^{3} – 360 = 216

â‡’Â a^{3} -b^{3} = 216 + 360 = 576

âˆ´ a^{3} – b^{3} = 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) (9x^{2} + 4y^{2} + 4z^{2} – 6xy – 4yz – 6zx)

(ii) (4x -3y + 2z) (16x^{2} + 9y^{2}+ 4z^{2} + 12xy + 6yz – 8zx)

(iii) (2a – 3b – 2c) (4a^{2} + 9b^{2} + 4c^{2} + 6ab – 6bc + 4ca)

(iv) (3x -4y + 5z) (9x^{2} + 16y^{2} + 25z^{2} + 12xy- 15zx + 20yz)

Solution:

(i) (3x + 2y + 2z) (9x^{2} + 4y^{2} + 4z^{2} – 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} Â + (2z)^{3} – 3 x 3x x 2y x 2z

= 27x^{3} + 8y^{3} + 8Z^{3} – 36xyz

(ii) (4x – 3y + 2z) (16x^{2} + 9y^{2} + 4z^{2} + 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}Â + (2z)^{3} – 3 x 4x x (-3y) x (2z)

= 64x^{3} – 27y^{3} + 8z^{3} + 72xyz

(iii) (2a -3b- 2c) (4a^{2} + 9b^{2} + 4c^{2} + 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)

= 8a^{3} – 21b^{3} -8c^{3} – 36abc

(iv) (3x – 4y + 5z) (9x^{2} + 16y^{2} + 25z^{2} + 12xy – 15zx + 20yz)

= [3x + (-4y) + 5z] [(3x)^{2} + (-4y)^{2} + (5z)^{2 }– 3x x (-4y) -(-4y) (5z) – 5z x 3x]

= (3x)^{3} + (-4y)^{3}Â + (5z)^{3} – 3 x 3x x (-4y) (5z)

= 27x^{3} – 64y^{3} + 125z^{3} + 180xyz

Question 2.

Evaluate:

Solution:

Question 3.

If x + y + z = 8 and xy + yz+ zx = 20, find the value of x^{3} + y^{3} + z^{3} – 3xyz.

Solution:

We know that

x^{3} + y^{3} + z^{3} – 3xyz = (x + y + z) (x^{2} + y^{2} + z^{2} -xy -yz – zx)

Now, x + y + z = 8

Squaring, we get

(x + y + z)^{2} = (8)^{2
}x^{2} + y^{2} + z^{2} + 2(xy + yz + zx) = 64

â‡’ x^{2} + y^{2} + z^{2} + 2 x 20 = 64

â‡’Â x^{2} + y^{2} + z^{2} + 40 = 64

â‡’Â x^{2} + y^{2} + z^{2} = 64 – 40 = 24

Now,

x^{3} + y^{3} + z^{3} – 3xyz = (x + y + z) [x^{2} + y^{2} + z^{2} – (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 a^{3} + b^{3} + c^{3} – 3abc.

Solution:

a + b + c = 9, ab + be + ca = 26

Squaring, we get

(a + b + c)^{2} = (9)^{2}

a^{2} + b^{2} + c^{2} + 2 (ab + be + ca) = 81

â‡’Â a^{2} + b^{2} + c^{2} + 2 x 26 = 81

â‡’ a^{2} + b^{2} + c^{2} + 52 = 81

âˆ´Â a^{2} + b^{2} + c^{2} = 81 – 52 = 29

Now, a^{3} + b^{3} + c^{3} – 3abc = (a + b + c) [(a^{2} + b^{2} + c^{2} – (ab + bc + ca)]

= 9[29 – 26]

= 9 x 3 = 27

Question 5.

If a + b + c = 9, and a^{2} + b^{2} + c^{2} = 35, find the value of a^{3} + b^{3} + c^{3} – 3abc.

Solution:

a + b + c = 9

Squaring, we get

(a + b + c)^{2} = (9)^{2
}â‡’Â a^{2} + b^{2} + c^{2} + 2 (ab + be + ca) = 81

â‡’Â 35 + 2(ab + bc + ca) = 81

2(ab + bc + ca) = 81 – 35 = 46

âˆ´Â ab + bc + ca = \(\frac { 46 }{ 2 }\) = 23

Now, a^{3} + b^{3} + c^{3} – 3abc

= (a + b + c) [a^{2} + b^{2} + c^{2} – (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 a^{2} + b^{2}.

Solution:

a + b = 7, ab = 12

Squaring both sides,

(a + b)^{2} = (7)^{2
}â‡’Â a^{2} + b^{2} + 2ab = 49

â‡’Â a^{2} + b^{2} + 2 x 12 = 49

â‡’ a^{2} + b^{2} + 24 = 49

â‡’ a^{2} + b^{2} = 49 – 24 = 25

âˆ´ a^{2} + b^{2} = 25

Question 4.

If a – b = 5 and ab = 12, find the value of a^{2} + b^{2}Â .

Solution:

a – b = 5, ab = 12

Squaring both sides,

â‡’ (a – b)^{2 } = (5)^{2
}â‡’Â a^{2} + b^{2} – 2ab = 25

â‡’Â a^{2} + b^{2} – 2 x 12 = 25

â‡’Â a^{2} + b^{2} – 24 = 25

â‡’Â a^{2} + b^{2} = 25 + 24 = 49

âˆ´ a^{2} + b^{2} = 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 andÂ ab + bc + ca = 23, then a^{2} + b^{2} + c^{2} =

(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
}a^{2} + b^{2} + c^{2} + 2 (ab + bc + ca) = 81

â‡’ a^{2} + b^{2} + c^{2} + 2 x 23 = 81

â‡’Â a^{2} + b^{2}+ c^{2} + 46 = 81

â‡’Â a^{2} + b^{2}+ c^{2} = 81 – 46 = 35Â Â (a)

Question 9.

(a – b)^{3} + (b – c)^{3} + (c – a)^{3} =

(a) (a + b + c) (a^{2} + b^{2} + c^{2} – 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}Â + (b- c)^{3} + (c- a)^{3
}âˆµ a – b + b – c + c – a =* 0
*âˆ´ (a

*–*b)

^{3}+ (b – c)

^{3}

*+ (c – a)*(a -b)(b- c) (c – a)Â Â Â Â (c)

^{3 }= 3Question 10.

Solution:

Question 11.

If a – b = -8 and ab = -12 then a^{3} – b^{3} =

(a) -244

(b) -240

(c) -224

(d) -260

Solution:

a – b = -8, ab = -12

(a – b)^{3} = a^{3} – b^{3} – 3ab (a – b)

(-8)^{3} = a^{3} – b^{3} – 3 x (-12) (-8)

-512 = a^{3}-b^{3}– 288

a^{3} – b^{3} = -512 + 288 = -224Â Â Â Â (c)

Question 12.

If the volume of a cuboid is 3x^{2} – 27, then its possible dimensions are

(a) 3, x^{2}, -27x

(b) 3, x – 3, x + 3

(c) 3, x^{2}, 27x

(d) 3, 3, 3

Solution:

Volume = 3x^{2} -27 = 3(x^{2} – 9)

= 3(x + 3) (x – 3)

âˆ´ Dimensions areÂ Â = 3, x – 3, xÂ Â +Â 3Â Â Â Â Â Â Â (b)

Question 13.

75 x 75 + 2 x 75 x 25Â Â Â + 25 x 25 is equal to

(a) 10000

(b) 6250

(c) 7500

(d) 3750

Solution:

Question 14.

(x – y) (x + y)(x^{2} + y^{2}) (x^{4} + y^{4}) is equal to

(a) x^{16} – y^{16}

(b) x^{8} – y^{8}

(c) x^{8} + y^{8}

(d) x^{16} + y^{16
}Solution:

Question 15.

Solution:

Question 16.

Solution:

Question 17.

Solution:

Question 18.

Solution:

Question 19.

If a^{2} + b^{2} + c^{2} – ab – bc – ca = 0, then

(a) a + b = c

(b) b + c = a

(c) c + a = b

(d) a = b = c

Solution:

a^{2} + b^{2} + c^{2} – ab – bc – ca = 0

2 {a^{2} + b^{2} + c^{2} – ab – be – ca) = 0 (Multiplying by 2)

â‡’Â 2a^{2} + 2b^{2} + 2c^{2}– 2ab – 2bc – 2ca = 0

â‡’Â a^{2} + b^{2} – 2ab + b^{2} + c^{2} – 2bc + c^{2} + a^{2} – 2ca = 0

â‡’Â (a – b)^{2} + (b – c)^{2} + (c – a)^{2} = 0

(a – b)^{2} = 0, then a – b = 0

â‡’ a = b

Similarly, (b – c)^{2} = 0, then

b-c = 0

â‡’ b = c

and (c – a)^{2} = 0, then c-a = 0

â‡’ c = a

âˆ´ a = b – cÂ Â Â Â Â Â Â Â Â Â (d)

Question 20.

Solution:

Question 21.

Solution:

Question 22.

If a + b + c = 9 and ab + bc + ca = 23, then a^{3} + b^{3} + c^{3} – 3 abc =

(a) 108

(b) 207

(c) 669

(d) 729

Solution:

a^{3} + b^{3} + c^{3} – 3abc

= (a + b + c) [a^{2} + b^{2} + c^{2} – (ab + bc + ca)

Now, a + b + c = 9

Squaring,

a^{2} + b^{2} + c^{2} + 2 (ab + be + ca) = 81

â‡’Â a^{2} + b^{2} + c^{2} + 2 x 23 = 81

â‡’Â a^{2} + b^{2} + c^{2} + 46 = 81

â‡’Â a^{2} + b^{2} + c^{2} = 81 – 46 = 35

Now, a^{3} + b^{3} + c^{3} – 3 abc = (a + b + c) [(a^{2 }+ b^{2} + c^{2}) – (ab + bc + ca)

= 9[35 -23] = 9 x 12= 108Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (a)

Question 23.

Solution:

Question 24.

The product (a + b) (a – b) (a^{2} – ab + b^{2}) (a^{2} + ab + b^{2}) is equal to

(a) a^{6} +Â Â b^{6}

(b) a^{6} – b^{6
}(c) a^{3} –Â b^{3}

(d) a^{3} + b^{3
}Solution:

(a + b) (a – b) (a^{2} – ab + b^{2}) (a^{2} + ab +b^{2})

= (a + b) (a^{2}-ab + b^{2}) (a-b) (a^{2} + ab + b^{2})

= (a^{3} + b^{3}) (a^{3} – b^{3})

= (a^{3})^{2} – (b^{3})^{2 }= a^{6} – b^{6}Â Â (b)

Question 25.

The product (x^{2} – 1) (x^{4} + x^{2} + 1) is equal to

(a) x^{8} –Â Â 1

(b) x^{8} + 1

(c) x^{6} –Â Â 1

(d) x^{6} + 1

Solution:

(x^{2} – 1) (x^{4} + x^{2} + 1)

= (x^{2})^{3} – (1)^{3} = x^{6} – 1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (c)

Question 26.

Solution:

Question 27.

Solution:

RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities

- RD Sharma Class 9th Solutions Chapter 4 Algebraic Identities Exercise 4.1

RD Sharma Class 9th Solutions Chapter 4 Algebraic Identities Exercise 4.2

RD Sharma Class 9th Solutions Chapter 4 Algebraic Identities Exercise 4.3 - RD Sharma Class 9th Solutions Chapter 4 Algebraic Identities Exercise 4.4
- RD Sharma Class 9th Solutions Chapter 4 Algebraic Identities Exercise 4.5