How can this expression be simplified? How do I factorize $$a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)?$$ 
I've tried it in different ways but failed. Wish some one could help solving it out.
 A: A simplification of André's answer:
The expression is a quadratic polynomial in $a$ that vanishes when $a=b$ or $a=c$.
The leading coefficient is clearly $b-c$.
Hence the expression is $(b-c)(a-b)(a-c)$.
A: Another proof. By Laplace expansion (along the last row):
$$ -\sum_{cyc}a^2(b-c) = \det\begin{pmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{pmatrix},\tag{1}$$
But the RHS is a Vandermonde matrix, whose determinant is well-known.
Gaussian elimination gives:
$$ D=\det\begin{pmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{pmatrix}=\det\begin{pmatrix} 1 & 0 & 0 \\ a & b-a & c-a \\ a^2 & b^2-a^2 & c^2-a^2 \end{pmatrix} \tag{2}$$
and by expanding along the first row and factoring out $(b-a)$ and $(c-a)$:
$$ D = (b-a)(c-a)\cdot \det\begin{pmatrix} 1 & 1 \\ b+a & c+a\end{pmatrix}=(b-a)(c-a)(c-b).\tag{3}$$
A: If we think of this as a polynomial in the variable $a$, it is $0$ if $a=b$, also if $a=c$. So $a-b$ and $a-c$ divide the polynomial. By symmetry $b-c$ divides the polynomial. So the polynomial should be equal to $(a-b)(a-c)(b-c)$ times a constant. In this case the constant is $1$.
A: A high school solution:
For this you have to break the symmetry between all variables. Explicitly, we'll write $\;c-a=(c-b)+(b-a)$. Thus
\begin{align*}
a^2(b-c)+b^2(c-a)+c^2(a-b)&=ab(a-b)+bc(b-c)+ca(c-a)\\
&=ab(a-b)+bc(b-c)+ca(c-b)+ca(b-a)\\
&=(a-b)(ab-ac)+(b-c)(bc-ac)\\
&=a(a-b)(b-c)+c(b-c)(b-a)\\
&=-(a-b)(b-c)(c-a)
\end{align*}
A: HINT:
I think you meant factorization 
$$a^2(b-c)+b^2(c-a)=-c(a^2-b^2)+ab(a-b)=?$$
