If, for a non-zero x, $5x^2+7x+5= 0$, then the value of $x^3 +\frac{1}{x^3}$ is:

-$\frac{182}{125}$

$5x^2+7x+5= 0$

Dividing by 5x on both sides.

If x + \(\frac{1}{x}\)  = n

then, $x^3 +\frac{1}{x^3}$ = n³ - 3n

x + \(\frac{1}{x}\) = \(\frac{7}{5}\)

$x^3 +\frac{1}{x^3}$ =  (\(\frac{7}{5}\))³ - 3 × \(\frac{7}{5}\) = -$\frac{182}{125}$