Solve the equations:

a) $x^{4}+x^{3}-3 a^{2} x^{2}-2 a^{2} x+2 a^{4}=0$

b) $x^{3}-3 x=a^{3}+a^{-3}$.
a) $x^{4}+x^{3}-3 a^{2} x^{2}-2 a^{2} x+2 a^{4}=x^{2}\left(x^{2}+x-a^{2}\right)-2 a^{2}\left(x^{2}+x-a^{2}\right)=\left(x^{2}-2 a^{2}\right)\left(x^{2}+x-a^{2}\right)$.

b) Let $b=a+a^{-1}$ and note that $b^{3}=\left(a^{3}+a^{-3}\right)+3\left(a+a^{-1}\right)$, which means $b^{3}-3 b=a^{3}+a^{-3}$. Thus, $b$ is one of the roots of our equation, and $x^{3}-3 x-\left(b^{3}-3 b\right)=(x-b)\left(x^{2}+b x+b^{2}-3\right)$. The discriminant of the equation $x^{2}+b x+b^{2}-3=0$ is

$b^{2}-4\left(b^{2}-3\right)=12-3 b^{2} \leq 0$, since $|b| \geq 2$. Therefore, this equation has a real root (equal to $-b / 2$) only when $a= \pm 1$.

## Answer

a) $x_{1,2}= \pm \sqrt{2} a, x_{3,4}=\frac{-1 \pm \sqrt{1+4 a^{2}}}{2}$.

b) For $a \neq 0, \pm 1 \quad x=a+a^{-1} ; \quad$ for $a=1 \quad x_{1}=2, x_{2}=-1 ; \quad$ for $a=-1 \quad x_{1}=-2, x_{2}=1$.