Proving determinant using properties of determinants $$\begin{vmatrix}
1 & 1 & 1\\ 
a & b & c\\ 
a^3 & b^3 & c^3 
\end{vmatrix} =
(a-b)(b-c)(c-a)(a+b+c)$$
we have to solve this by using the properties of determinants without actually expanding the determinant. I am Unable to think which calculation to apply so was hoping for an hint.
 A: Using $C_2'=C_2-C_1, C_3'=C_3-C_1$
$$\begin{vmatrix}
1 & 1 & 1\\ 
a & b & c\\ 
a^3 & b^3 & c^3 
\end{vmatrix}$$
$$=\begin{vmatrix}
1 & 0 & 0\\ 
a & b-a & c-a\\ 
a^3 & b^3-a^3 & c^3-a^3 
\end{vmatrix}$$
$$=-(a-b)(c-a)\begin{vmatrix}
1 & 0 & 0\\ 
a & 1 & 1\\ 
a^3 & b^2+ab+a^2 & c^2+ca+a^2 
\end{vmatrix}$$
Use $C_2'=C_2-C_3$
A: say,  $
\begin{vmatrix}
1 & 1 & 1\\ 
a & b & c\\ 
a^3 & b^3 & c^3 
\end{vmatrix} = f(a,b,c)
$
Clearly, when $a=b$, the first two columns are identical making $f = 0 $
So $(a-b)$ is a factor of $f$ by factor theorem
Similarly, $(b-c)$ and $(c-a)$ are also factors of $f$ :
$$f(a,b,c) =   (a-b)(b-c)(c-a)*g(a,b,c)$$
See if you can think of a way to find out $g(a,b,c)$
A: Apart from Ganeshie8's, these answers are longer than just expanding the determinant.  The conceptual answer is to say the determinant is an alternating degree 4 polynomial in $a,b,c$ with all coefficients 1.  So it is the Vandermonde polynomial on $a,b,c$ times the symmetric degree 1 polynomial in them with coefficients 1, which is $a+b+c$.  See http://en.wikipedia.org/wiki/Vandermonde_polynomial
A: HINT
Subtract 1st column from  second than 2nd from 3rd 
apply identity of a^3-b^3 and take a-b and b-c common from 1st two columns then subtract 1st from 2nd and then open it at last .
