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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.

Daniele Tampieri
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Sameer Nilkhan
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  • Nope it cant be. – aryan bansal Jan 19 '20 at 12:01
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    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
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    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

4 Answers4

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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 $a=b=c$

Sam
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  • Great post. Once you factored out $(a + b + c)$, how do you realize to further simplify the $(a^2 + b^2...)$ term? I had tried things like $(a(a - b) + b(b - c) + c(c-a))$ but these didn't get me anywhere. – SRobertJames Jun 19 '23 at 23:27
  • @SRobertJames It's not too intuitive but comes up naturally from solving lots of similar types of problems. – Sam Jul 03 '23 at 13:49
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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.

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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.

Oscar Lanzi
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Michael Rozenberg
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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:

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Will Jagy
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