## 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 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

⇒ a^{2} + b^{2} + c^{2} + 2 x 23 = 81

⇒ a^{2} + b^{2}+ c^{2} + 46 = 81

⇒ a^{2} + b^{2}+ c^{2} = 81 – 46 = 35 (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} + (b- c)^{3} + (c- a)^{3
}∵ a – b + b – c + c – a =* 0
*∴ (a

*–*b)

^{3}+ (b – c)

^{3}

*+ (c – a)*(a -b)(b- c) (c – a) (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 = -224 (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)

∴ 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)(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)

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

⇒ a^{2} + b^{2} – 2ab + b^{2} + c^{2} – 2bc + c^{2} + a^{2} – 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 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

⇒ a^{2} + b^{2} + c^{2} + 2 x 23 = 81

⇒ a^{2} + b^{2} + c^{2} + 46 = 81

⇒ 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= 108 (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} + b^{6}

(b) a^{6} – b^{6
}(c) a^{3} – 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} (b)

Question 25.

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

(a) x^{8} – 1

(b) x^{8} + 1

(c) x^{6} – 1

(d) x^{6} + 1

Solution:

(x^{2} – 1) (x^{4} + x^{2} + 1)

= (x^{2})^{3} – (1)^{3} = x^{6} – 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.

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