If $x^2 - 3\sqrt{2}x + 1= 0$, then what is the value of $x^3 + (\frac{1}{x^3})$ ?

$45\sqrt{2}$

We know that,

If x + \(\frac{1}{x}\)  = n

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

If $x^2 - 3\sqrt{2}x + 1= 0$,

then what is the value of $x^3 + (\frac{1}{x^3})$= ?

If $x^2 - 3\sqrt{2}x + 1= 0$,

Divide by x on the both sides of the equation,

x + \(\frac{1}{x}\) = $3\sqrt{2}$

Then, $x^3 +\frac{1}{x^3}$ = ($3\sqrt{2}$⁾³ - 3 × $3\sqrt{2}$ = $54\sqrt{2}$ - $9\sqrt{2}$ = $45\sqrt{2}$