7. Given $a, b \in \mathbf{R}^{+}$ and $\frac{\sin ^{4} x}{a}+\frac{\cos ^{4} x}{b}=\frac{1}{a+b}$, then $\frac{\sin ^{8} x}{a^{3}}+\frac{\cos ^{8} x}{b^{3}}=$ $\qquad$
7. $\frac{1}{(a+b)^{3}}$ Detailed explanation: By the Cauchy-Schwarz inequality, we get $\frac{\sin ^{4} x}{a}+\frac{\cos ^{4} x}{b}=\frac{\left(\sin ^{2} x\right)^{2}}{a}+\frac{\left(\cos ^{2} x\right)^{2}}{b} \geqslant \frac{\left(\sin ^{2} x+\cos ^{2} x\right)^{2}}{a+b}$ $=\frac{1}{a+b}$, with equality holding if and only if $\frac{\sin ^{2} x}{a}=\frac{\cos ^{2} x}{b}$. 

Combining $\sin ^{2} x+\cos ^{2} x=1$, we solve to get $\sin ^{2} x=\frac{a}{a+b}, \cos ^{2} x=\frac{b}{a+b}$, thus $\frac{\sin ^{8} x}{a^{3}}+\frac{\cos ^{8} x}{b^{3}}=\frac{1}{(a+b)^{3}}$.