8.176. $\sin ^{3} x(1+\operatorname{ctg} x)+\cos ^{3} x(1+\operatorname{tg} x)=2 \sqrt{\sin x \cos x}$.
Solution.

Domain of definition: $\left\{\begin{array}{l}\sin x \neq 0, \\ \cos x \neq 0, \\ \sin x \cos x>0 .\end{array}\right.$

Write the equation in the form

$$
\begin{aligned}
& \sin ^{3} x \cdot\left(1+\frac{\cos x}{\sin x}\right)+\cos ^{3} x \cdot\left(1+\frac{\sin x}{\cos x}\right)=2 \sqrt{\sin x \cos x} \Leftrightarrow \\
& \Leftrightarrow \frac{\sin ^{3} x(\sin x+\cos x)}{\sin x}+\frac{\cos ^{3} x(\cos x+\sin x)}{\cos x}-2 \sqrt{\sin x \cos x}=0 \Leftrightarrow \\
& \Leftrightarrow \sin ^{2} x(\sin x+\cos x)+\cos ^{2} x(\cos x+\sin x)-2 \sqrt{\sin x \cos x}=0 \Leftrightarrow \\
& \Leftrightarrow(\sin x+\cos x)\left(\sin ^{2} x+\cos ^{2} x\right)-2 \sqrt{\sin x \cos x}=0, \sin x- \\
& -2 \sqrt{\sin x \cos x}+\cos x=0, \sin x-2 \sqrt{\sin x \cos x}+\cos x=0 \\
& (\sqrt{\sin x}-\sqrt{\cos x})^{2}=0 \Leftrightarrow \sqrt{\sin x}-\sqrt{\cos x}=0
\end{aligned}
$$

Then $\sqrt{\sin x}=\sqrt{\cos x}$, or $\sin x=\cos x, \sin x>0, \cos x>0$. Dividing this equation by $\cos x \neq 0$, we get $\operatorname{tg} x=1, x=\frac{\pi}{4}+2 \pi k=\frac{\pi}{4}(8 k+1), k \in Z$.

Answer: $x=\frac{\pi}{4}(8 k+1), k \in Z$.