If $\frac{x^2+y^2}{x^2-y^2}=\frac{17}{8}$, then find the value of $\frac{x^3+y^3}{x^3-y^3}$

Given: $\frac{x^2+y^2}{x^2-y^2}=\frac{17}{8}$
As $\frac{x}{y}=\frac{5}{3}$
Cubing both sides, we get
$\frac{x^3}{y^3}=\frac{(5{)}^3}{(3{)}^3}=\frac{125}{27}$
Applying componendo and Dividendo
$\frac{x^3+y^3}{x^3-y^3}=\frac{125+27}{125-27}$
$\Rightarrow \frac{x^3+y^3}{x^3-y^3}=\frac{152}{98}$
$\Rightarrow \frac{x^3+y^3}{x^3-y^3}=\frac{76}{49}$