<s>[INST] 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
 [/INST] 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.
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