If $8x^2 + 9x + 8 = 0$, then the value of $x^3 + \frac{1}{x^3}$ is :

$\frac{999}{512}$

If x + \(\frac{1}{x}\)  = n

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

If $8x^2 + 9x + 8 = 0$

Now divide by 8x on both sides of the equation,

x + \(\frac{1}{x}\) = \(\frac{-9}{8}\) 

then, $x^3 +\frac{1}{x^3}$ = ( \(\frac{-9}{8}\) )³ - 3 × \(\frac{9}{8}\)

$x^3 +\frac{1}{x^3}$ = \(\frac{-729}{512}\) - \(\frac{27}{8}\)

$x^3 +\frac{1}{x^3}$ = \(\frac{-729 + 1728}{512}\)

$x^3 +\frac{1}{x^3}$ = $\frac{999}{512}$