$x^3-3x^2+(a^2+2)x-a^2$ 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)
 A: 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.
A: Using Prosthaphaeresis Formula,
$\sin\dfrac{2\pi x_1}3+\sin \dfrac{2\pi x_3}3=2\sin\dfrac{\pi(x_1+x_3)}3\cos\dfrac{\pi(x_1-x_3)}3$
As $x_1+x_2+x_3=3,$
$\sin\dfrac{\pi(x_1+x_3)}3=\sin\dfrac{\pi(3-x_2)}3=\sin\dfrac{\pi x_2}3$
So, we have $2\sin\dfrac{\pi x_2}3\left[\cos\dfrac{\pi(x_1-x_3)}3-2\cos\dfrac{\pi(x_2)}3\right]=0$
Now, $\cos\dfrac{\pi(x_2)}3=\cos\dfrac{\pi\{3-(x_1+x_3)\}}3=-\cos\dfrac{\pi(x_1+x_3)}3$
So, we have $2\sin\dfrac{\pi x_2}3\left[\cos\dfrac{\pi(x_1-x_3)}3+2\cos\dfrac{\pi(x_1+x_3)}3\right]=0$
As the product of two multiplicands is zero, at least one of them must be zero.
If $\sin\dfrac{\pi x_2}3=0,\dfrac{\pi x_2}3=n\pi\iff x_2=3n$ where $n$ is any integer
Else $\cos\dfrac{\pi(x_1-x_3)}3+2\cos\dfrac{\pi(x_1+x_3)}3=0$
$\iff3\cos\dfrac{\pi x_1}3\cos\dfrac{\pi x_3}3=\sin\dfrac{\pi x_1}3\sin\dfrac{\pi x_3}3$
$\iff\tan\dfrac{\pi x_1}3\tan\dfrac{\pi x_3}3=3$ which is not easily tractable
