# $a^3 + b^3 + c^3 = 3abc$ , can this be true only when $a+b+c = 0$ or $a=b=c$, or can it be true in any other case?

Since $$a^3 + b^3 + c ^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc ),$$ from this it is clear that if $$(a+b+c) = 0$$ , then $$a^3 + b^3 + c ^3 - 3abc = 0$$ and $$a^3 + b^3 + c ^3 = 3abc$$ . Also if $$a=b=c$$ , $$a^3 + b^3 + c^3 = a^3 + a^3 +a^3 = 3a^3 = 3aaa = 3abc$$ , hence $$a^3 + b^3 + c^3 = 3abc$$.
But I was wondering if $$a^3 + b^3 + c^3 = 3abc$$ can be true even when none of the above two relations are true. Please guide me toward a solution.

• Nope it cant be. – aryan bansal Jan 19 '20 at 12:01
• For $a=-2,b=e^{-2\pi i /3}, c=e^{-4\pi i /3}$ also holds, while these numbers are not all equal nor they sum $0$. They sum $-3$ instead. – MoonLightSyzygy Jan 19 '20 at 12:26
• If a,b,c are real then only possibility is what you have addressed i.e $a+b+c=0 \,or\,a=b=c$ but if $a,b,c\in C$ then $a\in R \,,\,b=a\omega \,,\,c=a\omega ^2$ are also one of set of solutions but this is also included in $a+b+c=0$. – mathsdiscussion.com Jan 19 '20 at 12:32
• For complex numbers $a$, $b$, and $c$, $a^3+b^3+c^3=3abc$ if and only if the (possibly degenerate) triangle formed by complex coordinates $a$, $b$, and $c$ is an equilateral, or has its centroid at the origin $0$ (or equivalently, $a+b+c=0$, $a+b\omega+c\omega^2=0$, or $a+b\omega^2+c\omega=0$, where $\omega$ is a complex root of $x^2+x+1=0$). From this observation, it follows immediately that when $a$, $b$, and $c$ are real, $a^3+b^3+c^3=3abc$ iff $a+b+c=0$ or $a=b=c$. – Batominovski Jan 19 '20 at 14:45

By your work: $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$$ Thus, $$a^3+b^3+c^3-3abc=0$$ for $$a+b+c=0$$ or for $$a^2+b^2+c^2-ab-ac-bc=0,$$ which is $$(a-b)^2+(a-c)^2+(b-c)^2=0,$$ which gives $$a=b=c.$$ Id est, we have no another cases for equality occurring for real values.

• There are comments involving complex solutions. I added a qualification that the proof holds in the real domain. – Oscar Lanzi Jan 19 '20 at 12:38
• Oscar, if in the given of the problem we don't say about complex numbers so we say about real numbers. It's a known rule. – Michael Rozenberg Jan 19 '20 at 13:26

Without loss of generality, suppose that $$a\ge b$$ and $$a\ge c$$. Then $$a^2+b^2+c^2-ab-bc-ca=(a-b)(a-c)+(b-c)^2\ge 0.$$

Equality can only occur if $$a=b=c$$ and so there are no other solutions.

The equation can also be factorized as follows- \begin{align}a^3+b^3+c^3-3abc&=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\\ &=\frac 12(a+b+c)(2a^2+2b^2+2c^2-2ab-2bc-2ca)\\ &=\frac 12(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2) \end{align} Hence the equation can only be true when either $$a+b+c=0$$ or $$a=b=c$$

The polynomial factors fully over the complex numbers, $$(a+b+c)(a+b \omega + c \omega^2)(a + b \omega^2 + c \omega) \; , \;$$ where $$\omega$$ is a primitive cube root of unity, either solution of $$x^2 + x + 1 =0$$

There is actually a concrete calculation that tells us whether a homogeneous cubic factors completely over the complexes. Here is an excerpt from an article by Brookfield: