If $x^2 - 6\sqrt{3} x + 1= 0, $ then the value of $x^3 + \frac{1}{x^3}$ will be :

$630\sqrt{3}$

We know that,

If x + \(\frac{1}{x}\)  = n

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

If $x^2 - 6\sqrt{3} x + 1= 0, $

then the value of $x^3 + \frac{1}{x^3}$

Divide by x on both the sides of If $x^2 - 6\sqrt{3} x + 1= 0, $ then we get,

x + \(\frac{1}{x}\) = $6\sqrt{3}$

then the value of $x^3 + \frac{1}{x^3}$ = ($6\sqrt{3}$)³ - 3 × $6\sqrt{3}$ = $648\sqrt{3}$ - $18\sqrt{3}$ = $630\sqrt{3}$