# Given $a+b+c=0$. Show that $a^3+b^3+c^3=3abc$. [closed]

Given $a+b+c=0$. Show that $a^3+b^3+c^3=3abc$

This is a problem 161 of The USSR olympiad problem book by Shklarsky, Chentzov and Yaglom.

The stupid, but effective, way is to write $c=-a-b$ and put that into $$a^3+b^3+c^3-3abc=a^3+b^3-(a+b)^3+3ab(a+b)=\cdots=0.$$

A more stylish way is to note that $a+b+c=0$ implies $$\pmatrix{a&b&c\\c&a&b\\b&c&a}\pmatrix{1\\1\\1}=\pmatrix{0\\0\\0}$$ which implies $$\det\pmatrix{a&b&c\\c&a&b\\b&c&a}=0.$$ Of course, $$\det\pmatrix{a&b&c\\c&a&b\\b&c&a}=a^3+b^3+c^3-3abc.$$

Let $a,b,c$ be the roots of the cubic equation $$x^3 - hx - k = 0$$

Then $$a^3+b^3+c^3 = (ha+k)+(hb+k)+(hc+k) = h(a+b+c)+3k = 3abc$$

incerting $$c=-a-b$$ then we have $$a^3+b^3+(-a-b)^3=-3ab(a+b)$$ and this is also the right Hand side. also note that $$a^3+b^3+c^3-3abc=\left( c+a+b \right) \left( {a}^{2}-ab-ac+{b}^{2}-bc+{c}^{2} \right)=0$$ since $$a+b+c=0$$

We can write $$a^3+b^3+c^3-3abc=(a+b+c)(a+b\omega+c{\omega}^2)(a+b{\omega}^2+c\omega)$$

$$a^3+b^3+c^3-3abc=a^3+3a^2b+3ab^2+b^3+c^3-3a^2b-3ab^2-3abc=$$ $$=(a+b)^3+c^3-3abc(a+b+c)=0$$

Rewrite: $$a+b+c=0 \Rightarrow a=-b-c$$ Cube it: $$a^3=-b^3-3b^2c-3bc^2-c^3 \Rightarrow a^3+b^3+c^3=3bc(-b-c)=3abc.$$