If $x^4 +\frac{1}{x^4}=1154$, where x > 0, than what is value of $x^3 +\frac{1}{x^3}$?
Answer & explanation
Correct answer: option 4
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