If $(x + \frac{1}{x})=\frac{11}{5}$, what is the value of $(x^3 + \frac{1}{x^3})$ ?

$4\frac{6}{125}$

We know that,

If x + \(\frac{1}{x}\)  = n

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

If $x + \frac{1}{x}=\frac{11}{5}$,

then, the value of $x^3 + \frac{1}{x^3}$ = ($\frac{11}{5}$)³ - 3 × $\frac{11}{5}$

= $\frac{1331}{125}$ - $\frac{33}{5}$

= $\frac{1331 - 825}{125}$

= $\frac{506}{125}$ = $4\frac{6}{125}$