CBSE Class 8 Mathematics Algebraic Expressions and Factorisation MCQs Set 27

Answer: (d) 2

Explanation:
1. \( x^3+\frac1{x^3}=2^3-3\times2=2 \).

Teacher's Note:
1. Recall the identity that fits the expression and substitute carefully.
2. Verify with a simple value of the variable to be sure.

Answer: (b) 10

Explanation:
1. \( \left(x-\frac1x\right)^2=102-2=100 \).
2. So \( x-\frac1x=10 \) (taking the positive root).

Teacher's Note:
1. Work step by step and check the result by substituting a convenient number.
2. Keep track of signs while expanding or factorising.

Answer: (a) 4

Explanation:
1. \( x^2+\frac1{x^2}=\sqrt{322+2}=18 \).
2. \( \left(x-\frac1x\right)^2=16 \), so \( x-\frac1x=4 \).

Teacher's Note:
1. Choose the standard identity or theorem first, then apply it.
2. Remainder and factor questions use the value that makes the divisor zero.

Answer: (a) 4

Explanation:
1. Let t = x + 1/x: t³ − 3t = 52.
2. t = 4 satisfies 64 − 12 = 52, and t³ − 3t − 52 = (t − 4)(t² + 4t + 13) has no other real root. So t = 4.

Teacher's Note:
1. Recall the identity that fits the expression and substitute carefully.
2. Verify with a simple value of the variable to be sure.

Answer: (d) 2

Explanation:
1. Let t = x − 1/x: t³ + 3t = 14.
2. t = 2 satisfies 8 + 6 = 14, and the cubic has no other real root. So t = 2.

Teacher's Note:
1. Work step by step and check the result by substituting a convenient number.
2. Keep track of signs while expanding or factorising.

Answer: (c) \( \frac{15}{4} \)

Explanation:
1. x + 1/x = 17/4 with integer x gives x = 4 (4 + 1/4 = 17/4).
2. Then x − 1/x = 4 − 1/4 = 15/4.

Teacher's Note:
1. Choose the standard identity or theorem first, then apply it.
2. Remainder and factor questions use the value that makes the divisor zero.

Answer: (b) 140

Explanation:
1. \( x^2+\frac1{x^2}=27 \) and \( \left(x-\frac1x\right)^2=25 \), so x − 1/x = 5.
2. \( x^3-\frac1{x^3}=5^3+3\times5=140 \).

Teacher's Note:
1. Recall the identity that fits the expression and substitute carefully.
2. Verify with a simple value of the variable to be sure.

Answer: (d) 322

Explanation:
1. \( x^3+\frac1{x^3}=27-9=18 \).
2. \( x^6+\frac1{x^6}=18^2-2=322 \).

Teacher's Note:
1. Work step by step and check the result by substituting a convenient number.
2. Keep track of signs while expanding or factorising.

Answer: (c) 52

Explanation:
1. Divide by t: t + 1/t = 4.
2. \( t^3+\frac1{t^3}=4^3-3\times4=52 \).

Teacher's Note:
1. Choose the standard identity or theorem first, then apply it.
2. Remainder and factor questions use the value that makes the divisor zero.

Answer: (a) 25

Explanation:
1. x² + y² = (x + y)² − 2xy = 49 − 24 = 25.

Teacher's Note:
1. Recall the identity that fits the expression and substitute carefully.
2. Verify with a simple value of the variable to be sure.

Answer: (b) \( (x+y+z)^3=27xyz \)

Explanation:
1. If a + b + c = 0 then a³ + b³ + c³ = 3abc, where a = x^(1/3) etc.
2. So x + y + z = 3(xyz)^(1/3), i.e. (x + y + z)³ = 27xyz.

Teacher's Note:
1. Work step by step and check the result by substituting a convenient number.
2. Keep track of signs while expanding or factorising.

Answer: (c) 3abc

Explanation:
1. a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca) = 0.
2. So a³ + b³ + c³ = 3abc.

Teacher's Note:
1. Choose the standard identity or theorem first, then apply it.
2. Remainder and factor questions use the value that makes the divisor zero.

Answer: (d) 3

Explanation:
1. Combine: \( \frac{a^3+b^3+c^3}{abc} \).
2. Since a + b + c = 0, a³ + b³ + c³ = 3abc, so the value is 3.

Teacher's Note:
1. Recall the identity that fits the expression and substitute carefully.
2. Verify with a simple value of the variable to be sure.

Answer: (a) 108

Explanation:
1. x² + y² + z² = 81 − 2 × 23 = 35.
2. x³ + y³ + z³ − 3xyz = (x + y + z)(x² + y² + z² − xy − yz − zx) = 9 × (35 − 23) = 108.

Teacher's Note:
1. Work step by step and check the result by substituting a convenient number.
2. Keep track of signs while expanding or factorising.

Answer: (d) 3

Explanation:
1. 5ˣ = 125 = 5³, so x = 3.

Teacher's Note:
1. Choose the standard identity or theorem first, then apply it.
2. Remainder and factor questions use the value that makes the divisor zero.

Answer: (c) 27

Explanation:
1. 3ˣ⁻¹(3 − 1) = 18, so 3ˣ⁻¹ = 9 = 3², giving x = 3.
2. xˣ = 3³ = 27.

Teacher's Note:
1. Recall the identity that fits the expression and substitute carefully.
2. Verify with a simple value of the variable to be sure.

Answer: (d) 25

Explanation:
1. 2ˣ⁻¹(2 − 1) = 16, so 2ˣ⁻¹ = 2⁴, giving x = 5.
2. x² = 25.

Teacher's Note:
1. Work step by step and check the result by substituting a convenient number.
2. Keep track of signs while expanding or factorising.

Answer: (c) \( \frac{4}{x-y} \)

Explanation:
1. Numerator = 4xy. Denominator = xy(x − y).
2. The fraction = \( \frac{4}{x-y} \).

Teacher's Note:
1. Choose the standard identity or theorem first, then apply it.
2. Remainder and factor questions use the value that makes the divisor zero.