<s>[INST] $x^3-3x^2+(a^2+2)x-a^2$ has 3 roots $x_1,x_2,x_3$ such that $\sin \tfrac{2\pi x_1}{3}+\sin \tfrac{2\pi x_3}{3}=2\sin \tfrac{2\pi x_2}{3}$. Find $a$. $x^3-3x^2+(a^2+2)x-a^2$ has 3 roots $x_1,x_2,x_3$ such that
$\sin \dfrac{2\pi x_1}{3}+\sin \dfrac{2\pi x_3}{3}=2\sin \dfrac{2\pi x_2}{3}$. 
Find $a$
(Bulgari 1998)
 [/INST] Observe
\begin{align*}x^3-3x^2+(a^2+2)x-a^2&=x(x^3-3x+2)+a^2(x-1)\\ &=[x(x-2)+a^2](x-1)\\ &=(x^2-2x+a^2)(x-1)\end{align*}
And the roots of $x^3-3x^2+(a^2+2)x-a^2$ are $1+\sqrt{1-a^2},\;1-\sqrt{1-a^2}\;\text{and}\; 1.$
If we set $a=1$ or $a=-1$, then the three roots are equal to $1$ and the equality holds.
</s>