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
 A: 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.
A: The first trick is to get as much zeroes as you can in the first row. That makes multiplication easier.
$\begin{vmatrix} 1 & a^2+bc & a^3 \\ 1 & b^2+ca & b^3\\ 1 & c^2+ab &c^3 \end{vmatrix}$
subtracting second row from first row and third row from second row: 
$(a-b)(b-c)\begin{vmatrix} 0 & a+b-c & a^2+ab+b^2 \\ 0 & b+c-a & b^2+bc+c^2\\ 1 & c^2+ab &c^3 \end{vmatrix}$
Subtracting second row from first row:
$-(a-b)(b-c)(c-a)\begin{vmatrix} 0 & 2 & a+b+c \\ 0 & b+c-a & b^2+bc+c^2\\ 1 & c^2+ab &c^3 \end{vmatrix}$
exchanging row and column: 
$(a-b)(b-c)(c-a)\begin{vmatrix} 0 & 0 & 1 \\ 2 & b+c-a & c^2+ab\\ a+b+c & b^2+bc+c^2 &c^3 \end{vmatrix}$
Now take determinant and get the result.
A: The solution-attempt in the question is excellent (after your edit)! 
But unfortunately you made a small mistake. After taking out the factor $(a-b)(c-a)$, the middle element should be $c-a-b$, not $a+b+c$. If you fix that, the remaining determinant can be expanded and easily factorised.
