Cubic roots of determinant.

Question

If x=a+2b satisfies the cubic (a,b element of R) f(x)= $$\left|\begin{matrix} a-x & b & b \\ b & a-x & b \\ b & b & a-x\end{matrix}\right|$$ =0, then it's other 2 roots are?

Answer 1

The roots of this determinant are the eigenvalues of a circulant matrix with row $[a \ b \ b ]$ and so are given by $a+b\omega^k + b\omega^{2k}$, where $\omega$ is a primitive cubic root of unity. Hence, they are $a+b+b=a+2b$ (for $k=0$) and $a+b(\omega+\omega^2)=a-b$ (for $k=1,2$).

Answer 2

$$f(x)=(a-x)^3+2b^3-3b^2(a-x)=0$$ putting $y=a-x$ you get $$y^3-3b^2 y+2b^3=0\ ,$$ you know that $y=-2b$ is a solution, so $$y^3-3b^2 y+2b^3=(y+2b)(y-b)^2\ .$$ The other solutions are thus $y=b$, i.e. $x=a-b$.