Proving determinant with variables I have a problem that asks:

Prove that det$\begin{pmatrix} 
1 &amp;&amp; 1 &amp;&amp; 1 \\ 
a &amp;&amp; b &amp;&amp; c \\
a^2 &amp;&amp; b^2 &amp;&amp; c^2 \end{pmatrix} = (c-a)(c-b)(b-a)$

I was thinking of solving it like normal and see if I can end up with $(c-a)(c-b)(b-a)$

What I did:

det$\begin{pmatrix}
b &amp;&amp; c \\
b^2 &amp;&amp; c^2 \end{pmatrix}$ - 
det$\begin{pmatrix}
a &amp;&amp; c \\
a^2 &amp;&amp; c^2 \end{pmatrix}$ +
det$\begin{pmatrix}
a &amp;&amp; b \\
a^2 &amp;&amp; b^2 \end{pmatrix}$

which gives 

$(bc^2 - cb^2)-(ac^2-ca^2)+(ab^2-ba^2)$

which simplifies to

$(b-a)c^2 + (a-c)b^2 + (c-b)a^2$ 

and this is where I got stuck. Am I even approaching this problem correctly? Any help would be appreciated.
which simplifies to $\;\; (b-a)c^2 + (a-c)b^2 + (c-b)a^2$


Hint: $\;(b-a)c^2 + (a-c)b^2 + (c\color{red}{-a+a}-b)a^2\,=\,\ldots$
If a=b, b=c, c=a then determinant value is 0 so
  (a-b)(b-c)(c-a) are factor of the determinant now comparing the leading coefficient so the Value of the determinant is (a-b)(b-c)(c-a)