$$\begin{vmatrix}x-a & a-b & b-c \\ c-a & x-b & a-c \\ b-a & c-b & x-c \end{vmatrix} =0 $$
I began expanding but quickly gave up, i couldn't factor it.Adding or subtracting rows/columns got me nowhere because i made no zeroes.
Any idea would be helpful. Hopefuly there is a more elegant solution than expansion and brute force,if that is the only way how to factor(apart from using cubical formula) .
Best Answer
Without constraints or additional information, your determinant is:
$$-a^3+a^2 (b+c+x)-a \left(b^2+2 b (c-x)-c^2+2 c x+x^2\right)+b^3-b^2 (c+x)+b \left(c^2+2 c x-x^2\right)-c^3+c^2 x-c x^2+x^3$$
and brute force is the only way forward. The three solutions (using the cubic equation) are:
$$\{x\to a-b+c\} \\ \left\{x\to b-\sqrt{-a^2+2 a c-c^2}\right\} \\ \left\{x\to \sqrt{-a^2+2 a c-c^2}+b\right\}$$