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I answered the first question by expansion $a^3+b^3+c^3=\frac12 (a+b+c) [(a-b)^2+(c-b)^2+(a-c)^2]$

Then the second asks about the nature of the triangle ABC for witch a b and c are the lengths and knowing that $a^3+b^3+c^3=3abc$. I tried substitution. But what comes after that $\frac12 (a+b+c) [(a-b)^2+(c-b)^2+(a-c)^2] - 3abc =0$?

user10354138
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user2161721
  • 709

3 Answers3

7

The OP nearly solved the problem. Just that his formula is a bit wrong.

\begin{align} a^3+b^3+c^3-3abc &=(a+b+c)(a^2+b^2+c^2-ab-bc-ca) \\ &= \frac{1}{2}(a+b+c)\left[(a-b)^2+(b-c)^2+(c-a)^2 \right] \end{align}

Since $a+b+c \gt 0$, \begin{align} a^3+b^3+c^3=3abc & \iff a^3+b^3+c^3-3abc = 0 \\ & \iff (a-b)^2+(b-c)^2+(c-a)^2=0 \\ & \iff a=b=c \end{align}

Therefore the triangle is equilateral iff $a^3+b^3+c^3=3abc.$

Li Kwok Keung
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5

It is clear that if the triangle is equilateral, the condition is verified.

The reciprocal property is also true.

Indeed, if $\omega:=e^{2 i \pi/3}$ (with property $\omega^3=1$ : it is a third root of unity), we have the complex factorization :

$$0=a^3+b^3+c^3-3abc=(a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega)\tag{*}$$

We can assume that $a=1$ without loss of generality (otherwise said, we take the first side as our unit of length). The first factor $a+b+c$ in (*) cannot be zero. Let us assume that it is the second factor which is zero :

$$1+b\omega+c\omega^2=0\tag{1}$$ As it is known that :

$$1+\omega+\omega^2=0\tag{2}$$

Subtracting (2) from (1), we get :

$$(b-1)\omega =-(c-1)\omega^2$$

which is equivalent to :

$$(b-1) =-(c-1)\omega\tag{3}$$

which is possible iff $$b=c=1$$

(indeed if $c\ne 1$ we would have in (3) a real number equal to a pure imaginary number).

Therefore $a=b=c=1$ and the triangle is equilateral.

Same deduction if it is the third factor in (*) which is zero.

Jean Marie
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Yet "an other" solution. The sides $a,b,c$ are positive numbers, and we have a given equality in inequality of the arithmetic and geometric means (AM and GM) of the numbers $a^3, b^3,c^3$: $$ \frac {a^3+b^3+c^3}3=\sqrt[3]{a^3\cdot b^3\cdot c^3}\ . $$ The equality holds if and only if the involved numbers (which are $a^3, b^3,c^3$ in the AM-GM inequality) are equal, so $a=b=c$, making the given triangle equilateral.

$\square$

dan_fulea
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