Show the value of determinant is 0 I dont have any idea how to show the value of the following determinant is 0 without expanding  the determinants
$$
    \begin{vmatrix}
    1 & a & a^2-bc \\
    1 & b & b^2-ca \\
    1 & c & c^2-ab \\
    \end{vmatrix}
$$
Any ideas?

$\Delta=\begin{vmatrix}
1 & a & a^2-bc \\
1 & b & b^2-ca \\
1 & c & c^2-ab \\
\end{vmatrix}=\begin{vmatrix}
1 & a & a^2 \\
1 & b & b^2 \\
1 & c & c^2\\
\end{vmatrix}-\begin{vmatrix}
1 & a & bc \\
1 & b & ca \\
1 & c & ab \\
\end{vmatrix}=\Delta_1-\Delta_2$, say.
Now, $\Delta_2=\begin{vmatrix}
1 & a & bc \\
1 & b & ca \\
1 & c & ab \\
\end{vmatrix}=\frac{1}{abc}\begin{vmatrix}
a & a^2 & abc \\
b & b^2 & abc \\
c & c^2 & abc \\
\end{vmatrix}$ $\quad(a,b,c\ne0)$
$\qquad\qquad=\begin{vmatrix}
a & a^2 & 1 \\
b & b^2 & 1 \\
c & c^2 & 1 \\
\end{vmatrix}=-\begin{vmatrix}
a & 1 & a^2 \\
b & 1 & b^2 \\
c & 1 & c^2 \\
\end{vmatrix}=\begin{vmatrix}
1 & a & a^2 \\
1 & b & b^2 \\
1 & c & c^2\\
\end{vmatrix}=\Delta_1$
The properties used are self-explanatory.

Alternatively, to show $\Delta_1=\Delta_2$ we can use the following argument:
Let $f(a,b,c)=\Delta_2-\Delta_1$
Then, $f(a,b,c)=k(a-b)(b-c)(a-c)$ for some constant $k$ as $f(b,b,c)=f(a,b,a)=f(a,b,b)=0$.
The coefficient of $a^2b$ in $f$ is $0\Rightarrow k=0.$
