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SSC CHSL 14 Nov 2025 Shift 2 Paper

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Question : 69 of 100

Marks: +1, -0

If

b-\frac{1}{b}

= 3 then calculate the value of

b^{6}+\frac{1}{b^{6}}

?

1298
1345
1987
1678

Solution:

Concept:

Use algebraic identities to find higher powers from the given difference. Repeated squaring and multiplication lead to the required expression.

Explanation:

Given

b - \frac{1}{b} = 3

.

Square both sides:

(b - \frac{1}{b})^2 = 9

gives

b^2 + \frac{1}{b^2} - 2 = 9

, so

b^2 + \frac{1}{b^2} = 11

.

Square again:

(b^2 + \frac{1}{b^2})^2 = 121

gives

b^4 + \frac{1}{b^4} + 2 = 121

, hence

b^4 + \frac{1}{b^4} = 119

.

Now multiply:

(b^2 + \frac{1}{b^2})(b^4 + \frac{1}{b^4}) = 11 \times 119 = 1309

.

Expand the product:

b^6 + \frac{1}{b^6} + b^2 + \frac{1}{b^2} = 1309

.

Substitute

b^2 + \frac{1}{b^2} = 11

:

b^6 + \frac{1}{b^6} + 11 = 1309

.

Therefore

b^6 + \frac{1}{b^6} = 1309 - 11 = 1298

.

Answer:

1298 (Option A)

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