Prove that the determinant is $(a-b)(b-c)(c-a)(a^2 + b^2 + c^2 )$ 
Prove that
  $$
        \begin{vmatrix}
        1 & a^2 + bc  & a^3 \\
        1 & b^2 + ac & b^3 \\
        1 & c^2 + ab & c^3 \\
        \end{vmatrix}
 =(a-b)(b-c)(c-a)(a^2 + b^2 + c^2 )$$


My work



I am getting - sign extra , please tell me why I am getting this
 A: It is faster to exploit Gaussian elimination:
$$\begin{eqnarray*} \det\begin{pmatrix}1 & a^2+bc & a^3 \\ 1 &b^2 +ac & b^3 \\ 1 & c^2+ab & c^3\end{pmatrix}&=&\det\begin{pmatrix}1 & a^2+bc & a^3 \\ 0 &(b-a)(a+b-c) & (b-a)(b^2+ba+a^2) \\ 0 & (c-a)(a+c-b) & (c-a)(c^2+ca+a^2)\end{pmatrix}\\[0.2cm]&=&(b-a)(c-a)\det\begin{pmatrix}1 & a^2+bc & a^3 \\ 0 &a+b-c & b^2+ba+a^2 \\ 0 & a+c-b & c^2+ca+a^2\end{pmatrix} \\[0.2cm]&=&(b-a)(c-a)\det\begin{pmatrix} a+b-c & b^2+ba+a^2 \\ a+c-b & c^2+ca+a^2\end{pmatrix}\\[0.2cm]&=&(b-a)(c-a)\det\begin{pmatrix} a+b-c & b^2+ba+a^2 \\ 2(c-b) & (c-b)(a+b+c)\end{pmatrix}\\[0.2cm]&=&(b-a)(c-a)(c-b)\det\begin{pmatrix} a+b-c & b^2+ba+a^2 \\ 2 & a+b+c\end{pmatrix}\end{eqnarray*}$$
and:
$$ (a+b-c)(a+b+c)-2(a^2+ab+b^2) = (a+b)^2-c^2-2(a^2+ab+b^2)=-(a^2+b^2+c^2).$$
An even faster approach is to notice that the determinant is clearly zero if $a=b,a=c$ or $b=c$. So the determinant is $(a-b)(b-c)(c-a)$ times a symmetric second-degree polynomial, that is not hard to find.
A: Alternatively, it is easily seen that the determinant of the given matrix is a homogeneous polynomial $P(a,b,c)$ in $a$, $b$, and $c$ of degree $5$.  Furthermore, $P(a,b,c)=0$ if any two of the three inputs are equal.  Therefore, $P(a,b,c)$ is divisible by $(b-c)(c-a)(a-b)$.  Hence, $$P(a,b,c)=(b-c)(c-a)(a-b)\,Q(a,b,c)$$ for some homogeneous quadratic polynomial $Q(a,b,c)$.  It is evident that $Q(a,b,c)$ is symmetric in $a,b,c$, whence $Q(a,b,c)=p\,\left(a^2+b^2+c^2\right)+q\,\left(bc+ca+ab\right)$ for some constants $p,q$.  As $$-4=P(-1,0,1)=(-1)(2)(-1)\,(2p-q)$$ and $$-30=P(-1,0,2)=(-2)(3)(-1)\,(5p-2q)\,,$$
we get $p=-1$ and $q=0$.  That is, $P(a,b,c)=-(b-c)(c-a)(a-b)\left(a^2+b^2+c^2\right)$.
Edit:  Just noticed that Jack D'Aurizio had mentioned this polynomial technique.
