Đẳng thức quen thuộc: \(a^3+b^3+c^3=3abc\Rightarrow\left[{}\begin{matrix}a=b=c\\a+b+c=0\end{matrix}\right.\)
Do \(a;b;c\) là 3 cạnh của tam giác nên \(a;b;c>0\Rightarrow a+b+c>0\)
Ta có:
\(a^3+b^3+c^3=3abc\Leftrightarrow a^3+3a^2b+3ab^2+b^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(\left(a+b\right)^2-\left(a+b\right)c+c^2\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-bc\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2-ab-ac-bc=0\) (do a+b+c>0)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\a-c=0\\b-c=0\end{matrix}\right.\) \(\Leftrightarrow a=b=c\)
Vậy ABC là tam giác đều
