If $x^4 +\frac{1}{x^4}=1154$, where x > 0, than what is value of $x^3 +\frac{1}{x^3}$?

198

If x⁴ + \(\frac{1}{x^4}\) = a

then x² + \(\frac{1}{x^2}\) = \(\sqrt {a + 2}\) = b

and x + \(\frac{1}{x}\) = \(\sqrt {b + 2}\)

According to the question,

$x^4 +\frac{1}{x^4}=1154$

then x² + \(\frac{1}{x^2}\) = \(\sqrt {1154 + 2}\) = 34

x + \(\frac{1}{x}\) = \(\sqrt {34 + 2}\) = 6

If x + \(\frac{1}{x}\)  = n

then, $x^3 +\frac{1}{x^3}$ = n³ - 3 × n

$x^3 +\frac{1}{x^3}$ = 6³ - 3 × 6 = 216 - 18 = 198