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

$-717\sqrt{2}$

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

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

then, $K^2+\frac{1}{K^2}$ = n² – 2 × k × \(\frac{1}{k}\)

If x + \(\frac{1}{x}\)  = n

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

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

then,$x^3 +\frac{1}{x^3}$ = ($-3\sqrt{2}$ )³ - 3 × $-3\sqrt{2}$, = $-45\sqrt{2}$

also, $x^2+\frac{1}{x^2}$ = ($-3\sqrt{2}$)² – 2

$x^2+\frac{1}{x^2}$ = 18 – 2 = 16

Put these values in x⁵ + $\frac{1}{x^5}$ =

x⁵ + $\frac{1}{x^5}$ = 16 × $-45\sqrt{2}$ - $-3\sqrt{2}$

x⁵ + $\frac{1}{x^5}$ = $-720\sqrt{2}$ + $3\sqrt{2}$  = $-715\sqrt{2}$