Is there an easy way of seeing that this sum is $0$?

Question

Let $a,b,c,d$ be real positive numbers. Is there an easy way of seeing that

$$(c-d)(b-d)(b-c)-(c-d)(a-d)(a-c)+(b-d)(a-d)(a-b)-(b-c)(a-c)(a-b)=0 $$

without doing the whole multiplication ?

Answer 1

Yes! Pick your favorite variable, say $a$, and treat the expression as a polynomial in $a$ (with coefficients that are themselves polynomials in $b,c,d$). It's easy to check that this polynomial has degree at most $2$; but it's also easy to check that it vanishes at $b$, $c$, and $d$ and thus has at least three roots. Therefore it must be the zero polynomial.

Answer 2

Write it as$$\begin{align}&\int_a^b(c-d)[2x-c-d]\mathrm{d}x+\int_c^d(a-b)[2y-a-b]\mathrm{d}y\\&=\int_a^b\mathrm{d}x\int_c^d\mathrm{d}y\{2x+2y-a-b-c-d\}.\end{align}$$The symmetry $x\to(a+b) /2-x,\,y\to(c+d)/2-y$ make the rest obvious.

Answer 3

Making two variables equal makes two products vanish and makes the remaining two cancel each other. E.g. with $a=b$, the equation turns to the identity

$$(c-d)(b-d)(b-c)-(c-d)(b-d)(b-c)\equiv 0.$$ Hence the given expression is of the form

$$P(a,b,c,d)(a-b)(a-c)(a-d)(b-c)(b-d)(c-d).$$

Because of the degrees, $P(a,b,c,d)\equiv 0$.