Mathematical Reflections 2012, Issue 5 - Problem J243

Problem:
Let $a,b,c$ be real numbers such that $$\left(-\dfrac{a}{2}+\dfrac{b}{3}+\dfrac{c}{6}\right)^3+\left(\dfrac{a}{3}+\dfrac{b}{6}-\dfrac{c}{2}\right)^3+\left(\dfrac{a}{6}-\dfrac{b}{2}+\dfrac{c}{3}\right)^3=\dfrac{1}{8}.$$ Prove that $$(a-3b+2c)(2a+b-3c)(-3a+2b+c)=9.$$

Proposed by Titu Andreescu.

Solution:
From the given equality, we have $$(-3a+2b+c)^3+(2a+b-3c)^3+(a-3b+2c)^3=27.$$
Let $x=-3a+2b+c, y=2a+b-3c, z=a-3b+2c$. Since $x+y+z=0$, using the well known identity
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx),$$ we get $x^3+y^3+z^3=3xyz$, i.e.
$$3(-3a+2b+c)(2a+b-3c)(a-3b+2c)=27,$$ which gives $(-3a+2b+c)(2a+b-3c)(a-3b+2c)=9$.