If $x^2 -\sqrt{7}x +1 =0$, then $(x^3 + x^{-3}) = ?$

$4\sqrt{7}$

If x + \(\frac{1}{x}\)  = n

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

If $x^2 -\sqrt{7}x +1 =0$,

then $(x^3 + x^{-3}) = ?$

divide by x on both sides in $x^2 -\sqrt{7}x +1 =0$

x + \(\frac{1}{x}\) = $\sqrt{7}$

then $(x^3 + x^{-3})$ = ($\sqrt{7}^3$) - 3 × $\sqrt{7}$

$(x^3 + x^{-3})$ = 7$\sqrt{7}$ - 3$\sqrt{7}$ = $4\sqrt{7}$