$8.453 \tan^{3} x+\tan^{2} x+\cot^{2} x+\cot^{3} x-4=0$.
## Solution.

$\left(\operatorname{tg}^{3} x+\operatorname{ctg}^{3} x\right)+\left(\operatorname{tg}^{2} x+\operatorname{ctg}^{2} x+2\right)-6=0 \Leftrightarrow$

$\Leftrightarrow(\operatorname{tg} x+\operatorname{ctg} x)\left(\operatorname{tg}^{2} x+\operatorname{ctg}^{2} x-1\right)+(\operatorname{tg} x+\operatorname{ctg} x)^{2}-6=0 \Leftrightarrow$

$\Leftrightarrow(\operatorname{tg} x+\operatorname{ctg} x)\left((\operatorname{tg} x+\operatorname{ctg} x)^{2}-3\right)+(\operatorname{tg} x+\operatorname{ctg} x)^{2}-6=0 \Leftrightarrow$

Let $y=\operatorname{tg} x+\operatorname{ctg} x$.

Then $y^{3}+y^{2}-3 y-6=0 \Leftrightarrow(y-2)\left(y^{2}+3 y+3\right)=0 \Leftrightarrow y=2 \Leftrightarrow$ $\Leftrightarrow \operatorname{tg} x+\operatorname{ctg} x-2=0 \Leftrightarrow(\operatorname{tg} x-1)^{2}=0 \Leftrightarrow x=\frac{\pi}{4}+\pi k$.

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