# 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.

• As a start: If you take $a=b$ then the expression becomes $0$, and so $(a-b)$ is a factor of the expression. Similarly, $(b-c)$ and $(c-a)$ are also factors. Commented Sep 4, 2015 at 14:40

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 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*}

• $c-a = (c-b)+(b-a)$ Commented Sep 4, 2015 at 14:44
• I've just seen that. Thanks for pointing the typo! Commented Sep 4, 2015 at 15:20

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 $(b-c)(a-b)(a-c)$.

• An improvement! My initial writeup kept full symmetry to the end ($a-b$ divides, so does $b-c$, so does $c-a$, constant is $-1$). Then I partly broke symmetry, but breaking a little more is better. Commented Sep 4, 2015 at 14:56

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 $(c-a)$: $$D = (b-a)(c-a)\cdot \det\begin{pmatrix} 1 & 1 \\ b+a & c+a\end{pmatrix}=(b-a)(c-a)(c-b).\tag{3}$$

HINT:

I think you meant factorization

$$a^2(b-c)+b^2(c-a)=-c(a^2-b^2)+ab(a-b)=?$$

• Yeah. I meant factorization. Commented Sep 4, 2015 at 14:33