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

18

$ 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