lean_workbook_plus_30747

$a^{2}+b^{2}+c^{2}=ab+bc+ca$ yields $\left(a^{2}+b^{2}+c^{2}\right)-\left(ab+bc+ca\right)=0$ . But some algebra shows that $\left(c-a\right)^{3}-\left(a-b\right)^{3}=\left(\left(c-a\right)-\left(a-b\right)\right)\cdot\left(\left(a^{2}+b^{2}+c^{2}\right)-\left(ab+bc+ca\right)\right) =\left(\left(c-a\right)-\left(a-b\right)\right)\cdot 0=0$ , so that $\left(c-a\right)^{3}=\left(a-b\right)^{3}$ . Since $x^{3}=y^{3}\Rightarrow x=y$ , this yields $c-a=a-b$ , and similarly we find that $a-b=b-c$ . Thus, the trivial equation $\left(c-a\right)+\left(a-b\right)+\left(b-c\right)=0$ becomes $\left(a-b\right)+\left(a-b\right)+\left(a-b\right)=0$ , or, equivalently, $\left(1+1+1\right)\left(a-b\right)=0$ . Since $1+1+1$ is invertible, this yields $a-b=0$ , so that $a=b$ . Similarly, $b=c$ . Thus, $a=b=c$ , and we are done.