Proving determinants using properties of determinants $$\begin{vmatrix}
1 & a^2+bc  & a^3\\
1 & b^2+ca  & b^3\\
1 & c^2+ab  & c^3 
\end{vmatrix}
 =  (a-b)(b-c)(c-a)(a^2+b^2+c^2)$$
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.
Edit: just tried the problem and here is how I have done it
$$\begin{vmatrix}
1 & a^2+bc  & a^3\\
1 & -(a^2+b^2)+c(a-b)  & -(a^3-b^3)\\
1 & c^2-a^2-b(c-a)  & c^3-a^3 
\end{vmatrix}$$
then
$$\begin{vmatrix}
1 & a^2+bc  & a^3\\
0 & -((a-b)(a+b))+c(a-b)  & -((a-b)(a^2+ab+b^2))\\
0 & (c-a)(c+a)-b(c-a)  & (c-a)(c^2+ca+a^2) 
\end{vmatrix}$$
then
$$(a-b)(c-a)\begin{vmatrix}
1 & a^2+bc  & a^3\\
0 & a+b+c  & -(a^2+ab+b^2)\\
0 & a-b+c  & (c^2+ca+a^2) 
\end{vmatrix}$$
then
$$(a-b)(c-a)\begin{vmatrix}
  a+b+c  & -(a^2+ab+b^2)\\
  a-b+c  & (c^2+ca+a^2) 
\end{vmatrix}$$
then 
$$(a-b)(c-a) *  (  (a+b+c)(c^2+ca+a^2)-(a-b+c)(-(a^2+ab+b^2))  )$$
Here I am Confused on how to multiply them and get the answer
note: typed because I own a very bad handwriting 
Thank you every one for your help

Seeing the column of $1$'s, a first thought would be to subtract the first row from the other two. This turns out to give you a factor $b-a$ in the second row and a factor $c-a$ in the third.
$$
\begin{align}
\begin{vmatrix}
1 & a^2+bc  & a^3\\
1 & b^2+ca  & b^3\\
1 & c^2+ab  & c^3 
\end{vmatrix}
&=
\begin{vmatrix}
1 & a^2+bc  & a^3\\
0 & b^2-a^2+c(a-b)  & b^3-a^3\\
0 & c^2-a^2+b(a-c)  & c^3-a^3 
\end{vmatrix}
\\{}&
=(b-a)(c-a)
\begin{vmatrix}
1 & a^2+bc  & a^3\\
0 & (b+a)-c  & a^2+ab+b^2\\
0 & (c+a)-b  & a^2+ac+c^2 
\end{vmatrix}
\end{align}
$$
Now you can subtract the second row from the third, which makes the latter divisible by $b-c$, so you can factor again. I shouldn't be too difficult to complete the computation.
