3. Solve the inequality

$$
\frac{(|x+1|-|x-1|)\left(x^{3}-7 x^{2}+36\right)}{x^{8}+2 x^{6}-6 x^{4}+2 x^{2}+1} \geq 0
$$
Solution. Let's regroup the terms in the denominator:

$$
x^{8}+2 x^{6}-6 x^{4}+2 x^{2}+1=x^{8}-2 x^{4}+1+2 x^{6}-4 x^{4}+2 x^{2}=\left(x^{4}-1\right)^{2}+2 x^{2}\left(x^{2}-1\right)^{2} \geq 0
$$

Multiply the fraction by a positive quantity, and we get $\frac{\left(|x+1|^{2}-|x-1|^{2}\right)(x+2)(x-3)(x-6)}{\left(\left(x^{2}+1\right)^{2}+2 x^{2}\right)\left(x^{2}-1\right)^{2}(|x+1|+|x-1|)} \geq 0 \Rightarrow \frac{(4 x)(x+2)(x-3)(x-6)}{\left(x^{2}-1\right)^{2}} \geq 0$

Answer: $(-\infty ;-2] \cup[0 ; 1) \cup(1 ; 3] \cup[6 ;+\infty)$.