5. Note that $x=0$ is not a root of the equation for any $a$. Transform the equation by dividing it by $x^{3}$:

$$
\left(a^{2}+1\right)\left(x+\frac{1}{x}\right)^{3}-3\left(x+\frac{1}{x}\right)+2=12 a
$$

Thus, if $x_{0}$ is a root of the equation, then $\frac{1}{x_{0}}$ is also a root. Therefore, the only possible roots are $x=1$ or $x=-1$.

In the first case, we get $8 a^{2}+4=12 a$, from which $a=1$ or $a=\frac{1}{2}$, and in the second case, we have $8 a^{2}=-12 a$, from which $a=0$ or $a=-\frac{3}{2}$. Substituting the values $a=-3 / 2,1 / 2,1$ into the equation $\left(a^{2}+1\right) t^{3}-3 t+2-12 a=0$, where $t=x+\frac{1}{x}$, we verify that for each of them, it has a unique solution $t=2$ or $t=-2$, and the original equation has a unique solution $x=1$ or $x=-1$. In the case $a=0$, the equation $t^{3}-3 t+2=0$ has two solutions $t=-2$ and $t=1$. The root $t=-2$ gives the solution $x=-1$, while the root $t=1$ does not provide solutions for $x$.
Answer: $a=-\frac{3}{2}, 0, \frac{1}{2}, 1$.

Answer to option: $5-2: a=-\frac{2}{3}, 1, 2$.

## Lomonosov Moscow State University

Olympiad "Conquer Sparrow Hills"

Variant 2-1 (Chelyabinsk)