If $x + \frac{1}{x} = -2\sqrt{3}$, what is the value of $x^5 +\frac{1}{x^5}$ ?

$-178\sqrt{3}$

We know that,

 x⁵ + $\frac{1}{x^5}$ = (x² + $\frac{1}{x^2}$) × (x³ + $\frac{1}{x^3}$) – (x + $\frac{1}{x}$)

We also know that,

If $K+\frac{1}{K}=n$

then, $K^2+\frac{1}{K^2}$ = n² – 2

If x + \(\frac{1}{x}\)  = n

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

So,

If $x + \frac{1}{x} = -2\sqrt{3}$,

Then the value of (x² + $\frac{1}{x^2}$) = ($-2\sqrt{3}$)² – 2 = 10

and, $x^3 +\frac{1}{x^3}$ = ($-2\sqrt{3}$)³ - 3 × $-2\sqrt{3}$ = $-24\sqrt{3}$ + $6\sqrt{3}$ = $-18\sqrt{3}$

what is the value of $x^5 +\frac{1}{x^5}$ = ($-18\sqrt{3}$) (10) - ($-2\sqrt{3}$)= $-178\sqrt{3}$