4B. In the set of real numbers, solve the equation:

$$
\left|x^{4}-x^{2}-6\right|=\left|x^{4}-4\right|-\left|x^{2}+2\right|
$$
Solution. The left side of the equation can be written in the form

$$
\begin{aligned}
\left|x^{4}-x^{2}-6\right| & =\left|x^{4}-4-\left(x^{2}+2\right)\right|=\left|\left(x^{2}+2\right)\left(x^{2}-2\right)-\left(x^{2}+2\right)\right| \\
& =\left(x^{2}+2\right)\left|x^{2}-2-1\right|=\left(x^{2}+2\right)\left|x^{2}-3\right|
\end{aligned}
$$

and the right side can be written in the form

$$
\left|x^{4}-4\right|-\left|x^{2}+2\right|=\left|\left(x^{2}+2\right)\left(x^{2}-2\right)\right|-\left|x^{2}+2\right|=\left(x^{2}+2\right)\left(\left|x^{2}-2\right|-1\right)
$$

In $(*)$ and $(* *)$, we used the fact that $x^{2}+2>0$ for every $x \in \mathbb{R}$. The initial equation can be written in the form

$$
\left(x^{2}+2\right)\left|x^{2}-3\right|=\left(x^{2}+2\right)\left(\left|x^{2}-2\right|-1\right)
$$

and if we divide by $x^{2}+2$, it takes the form

$$
\left|x^{2}-3\right|=\left(\left|x^{2}-2\right|-1\right)
$$

Now we will consider three cases.

a) If $|x| \geq \sqrt{3}$, the equation becomes the identity $x^{2}-3=x^{2}-2-1=x^{2}-3$.

b) If $\sqrt{2} \leq x<\sqrt{3}$, the equation becomes the equation $3-x^{2}=x^{2}-2-1$ whose solution is $x=\sqrt{3}$, which does not belong to the interval $\sqrt{2} \leq x<\sqrt{3}$.

c) If $|x|<\sqrt{2}$, the equation becomes $3-x^{2}=2-x^{2}-1$ which is not possible.

Therefore, the solution to the equation is every $x,|x| \geq \sqrt{3}$.