$$
\begin{array}{l}
\text { 5. Let } \frac{\sin ^{4} \theta}{a}+\frac{\cos ^{4} \theta}{b}=\frac{1}{a+b}(a, b>0) \text {. } \\
\text { Then } \frac{\sin ^{8} \theta}{a^{3}}+\frac{\cos ^{8} \theta}{b^{3}}=
\end{array}
$$
5. $\frac{1}{(a+b)^{3}}$.

Let $\sin ^{2} \theta=x$. Then the given equation becomes $\frac{x^{2}}{a}+\frac{(1-x)^{2}}{b}=\frac{1}{a+b}$.
Solving for $x$, we get $x=\frac{a}{a+b}, 1-x=\frac{b}{a+b}$.
Thus, $\frac{\sin ^{8} \theta}{a^{3}}+\frac{\cos ^{8} \theta}{b^{3}}=\frac{x^{4}}{a^{3}}+\frac{(1-x)^{4}}{b^{3}}$
$=\frac{1}{(a+b)^{4}}(a+b)=\frac{1}{(a+b)^{3}}$.