If \({k^4} + \frac{1}{{{k^4}}} = 47\), then what is the value of \({k^3} + \frac{1}{{{k^3}}}\)?

Question

Question

This question was previously asked in

SSC CGL 2022 Tier-I Official Paper (Held On : 05 Dec 2022 Shift 1)

Answer 1

Given:

\({k^4} + \frac{1}{{{k^4}}} = 47\)

Concept used:

If \(\rm {a} + \frac{1}{{{a}}} = b\)

The,

\(\rm {a^3} + \frac{1}{{{a^3}}} = b^3-3b\)

\(\rm ({a} + \frac{1}{{{a}}})^2 = a^2 + \frac{1}{a^2}+2\)

Calculation:

\({k^4} + \frac{1}{{{k^4}}} = 47\)

⇒ \({k^4} + \frac{1}{{{k^4}}} +2= 47+2\)

⇒ \(({k^2} + \frac{1}{{{k^2}}})^2= 49\)

⇒ \({k^2} + \frac{1}{{{k^2}}}= 7\)

⇒ \({k^2} + \frac{1}{{{k^2}}}+2= 7+2\)

⇒ \(({k} + \frac{1}{{{k}}})^2= 9\)

⇒ \({k} + \frac{1}{{{k}}}= 3\)

Now,

\({k^3} + \frac{1}{{{k^3}}}\) = 3³ - 3 × 3

⇒ \({k^3} + \frac{1}{{{k^3}}}\) = 27 - 9

⇒ \({k^3} + \frac{1}{{{k^3}}}\) = 18

∴ The required answer is 18.

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