3.288. $\operatorname{tg}^{3} x-\operatorname{tg}^{2} x-3 \operatorname{tg} x+3$.

3.288. $\tan^{3} x-\tan^{2} x-3 \tan x+3$.
Solution.

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

$=\frac{\frac{2}{\sqrt{2}}\left(\frac{\sqrt{2}}{2} \sin x-\frac{\sqrt{2}}{2} \cos x\right) \cdot 2\left(\frac{1}{2} \sin x-\frac{\sqrt{3}}{2} \cos x\right) \cdot 2\left(\frac{1}{2} \sin x+\frac{\sqrt{3}}{2} \cos x\right)}{\cos ^{3} x}=$

$=\frac{4 \sqrt{2}\left(\cos 45^{\circ} \sin x-\sin 45^{\circ} \cos x\right)\left(\cos 60^{\circ} \sin x-\sin 60^{\circ} \cos x\right)\left(\cos 60^{\circ} \sin x+\sin 60^{\circ} \cos x\right)}{\cos ^{3} x}=$

$=\frac{4 \sqrt{2}\left(\sin x \cos 45^{\circ}-\cos x \sin 45^{\circ}\right)\left(\sin x \cos 60^{\circ}-\cos x \sin 60^{\circ}\right)\left(\sin x \cos 60^{\circ}+\cos x \sin 60^{\circ}\right)}{\cos ^{3} x}=$

$=[\sin \alpha \cos \beta \pm \cos \alpha \sin \beta=\sin (\alpha \pm \beta)]=$

$=\frac{4 \sqrt{2} \sin \left(x-45^{\circ}\right) \sin \left(x-60^{\circ}\right) \sin \left(x+60^{\circ}\right)}{\cos ^{3} x}$.

Answer: $\frac{4 \sqrt{2} \sin \left(x-45^{\circ}\right) \sin \left(x-60^{\circ}\right) \sin \left(x+60^{\circ}\right)}{\cos ^{3} x}$.