If  $\frac{{\sin }^4\theta }{a}+\frac{{\cos }^4\theta }{b}=\frac{1}{a+b}then\frac{{\sin }^8\theta }{a^3}+\frac{{\cos }^8\theta }{b^3}=\_\_\_$

$$\begin{gathered} \frac{{\sin }^4\theta }{a}+\frac{{(1-{\sin }^2\theta )}^2}{b}=\frac{1}{a+b} \\ {(a+b)}^2{\sin }^4\theta -2a(a+b){\sin }^2\theta +a^2=0 \\ (a+b){\sin }^2\theta =a\Rightarrow {\sin }^2\theta =\frac{a}{a+b} \\ {\cos }^2\theta =\frac{b}{a+b} \\ Now\frac{{\sin }^8\theta }{a^3}+\frac{{\cos }^8\theta }{b^3}=\frac{1}{a^3}{\left(\frac{a}{a+b}\right)}^4+\frac{1}{b^3}{\left(\frac{b}{a+b}\right)}^4=\frac{1}{{(a+b)}^3} \end{gathered}$$