Linear Algebra determinant reduction Prove, without expanding, that
\begin{vmatrix}
1 &a  &a^2-bc \\ 
1 &b  &b^2-ca \\ 
1 &c  &c^2-ab 
\end{vmatrix} vanishes.
Any hints ?
 A: Let
$$f(a)=\begin{vmatrix}
1 &a  &a^2-bc \\ 
1 &b  &b^2-ca \\ 
1 &c  &c^2-ab 
\end{vmatrix}$$
then it's easy to see that $f$ is a polynomial on $a$ with degree at most $2$ and $f(b)=f(c)=0$ so
$$f(a)=\lambda(a-b)(a-c)$$
now, WLOG assume that $bc\ne0$, take $a=0$; we see easily that $f(0)=0$ hence $\lambda=0$.
A: \begin{align}
  \begin{vmatrix}
    1 & a & a^2-bc \\ 
    1 & b & b^2-ca \\ 
    1 & c & c^2-ab 
  \end{vmatrix}
&=\begin{vmatrix}
    1 & a    & a^2-bc        \\ 
    0 & b-a  & b^2-a^2+bc-ca \\ 
    0 & c-b  & c^2-b^2+ca-ab 
  \end{vmatrix} \\
&=\begin{vmatrix}
    b-a & b^2-a^2+bc-ca \\ 
    c-b & c^2-b^2+ca-ab 
  \end{vmatrix} \\
&=\begin{vmatrix}
    b-a & (b-a)(b+a+c) \\ 
    c-b & (c-b)(c+b+a) 
  \end{vmatrix}
\end{align}
A: e.g. if your matrix is $A$, consider $(b-c,c-a,a-b) A$
A: Add $(ab+bc+ca)$ times the first column to the last and find the common factor of the last column.
A: Simply add a multiple of the first column to the last
$$\begin{vmatrix} 1 & a & a^2-bc\\1&b&b^2-ac\\1&c&c^2-ab\end{vmatrix}
=\begin{vmatrix} 1 & a & a^2-bc+(1)(ab+ac+bc)\\1&b&b^2-ac+(1)(ab+ac+bc)\\1&c&c^2-ab+(1)(ab+ac+bc)\end{vmatrix}
= \begin{vmatrix} 1 & a & a(a+b+c)\\1&b&b(a+b+c)\\1&c&c(a+b+c)\end{vmatrix}
= 0$$
The final step follows because the second and third column are linearly dependent.
