IF $ k^4+\frac{1}{k^4} = 47$, then what is the value of $ k^3 +\frac{1}{k^3}$ ?
Answer & explanation
Correct answer: option 3
$ k^4+\frac{1}{k^4} = 47$
what is the value of $ k^3 +\frac{1}{k^3}$
If x4 + \(\frac{1}{x^4}\) = a
then x2 + \(\frac{1}{x^2}\) = \(\sqrt {a + 2}\) = b
and x + \(\frac{1}{x}\) = \(\sqrt {b + 2}\)
If $ k^4+\frac{1}{k^4} = 47$
then, $ k^2+\frac{1}{k^2}$ = \(\sqrt {47 + 2}\) = 7
and, k + \(\frac{1}{k}\) = \(\sqrt {7 + 2}\) = 3
$ k^3 +\frac{1}{k^3}$ = 33 - 3 × 3 = 18