## RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3

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
}⇒ a^{3} + 6^{3} + 3ab (a + b) = 1000

⇒ a^{3} + b^{3} + 3 x 21 x 10 = 1000

⇒ a^{3} + b^{3} + 630 = 1000

⇒ a^{3} + b^{3} = 1000 – 630 = 370

∴ 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,

⇒ (a – A)^{3} = (4)^{3
}⇒ a^{3} – b^{3} – 3ab (a – b) = 64

⇒ a^{3}-i^{3}-3×21 x4 = 64

⇒ a^{3} – 6^{3} – 252 = 64

⇒ a^{3} – 6^{3} = 64 + 252 =316

∴ 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
}⇒ (2x)^{3} + (3y)^{3} + 3 x 2x x 3X2x + 3y) = 2197

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

⇒ 8x^{3} + 27y^{3} + 18 x 6 x 13 = 2197

⇒ 8X^{3} + 27y^{3} + 1404 = 2197

⇒ 8x^{3} + 27y^{3} = 2197 – 1404 = 793

∴ 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
}⇒ (3x)^{3} – (2y)^{3} – 3 x 3x x 2y(3x – 2y) =1331

⇒ 27x^{3} – 8y^{3} – 18xy(3x -2y) =1331

⇒ 27x^{3} – 8y^{3} – 18 x 12 x 11 = 1331

⇒ 27x^{3} – 8y^{3} – 2376 = 1331

⇒ 27X^{3} – 8y^{3} = 1331 + 2376 = 3707

∴ 2x^{3} – 8y^{3} = 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 = a^{3} + b^{3} + 3ab(a + b) and (a – b)^{3}= a^{3} – b^{3} – 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} = 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) 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
}⇒ (4x)^{3} – (5y)^{3} – 3 x 4x x 5z(4x – 5z) = 4096

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

⇒ 64x^{3} – 125z^{3} – 60 x 12 x 16 = 4096

⇒ 64x^{3} – 125z^{3} – 11520 = 4096

⇒ 64x^{3} – 125z^{3} = 4096 + 11520 = 15616

Question 16.

Solution:

Question 17.

Simplify each of the following:

Solution:

Question 18.

Solution:

Question 19.

Solution:

Hope given RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 are helpful to complete your math homework.

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