TITLE: Showing an equality of complex numbers
QUESTION [6 upvotes]: Let $e_k \in \mathbb{C}$ such that
$$e_1+e_2+e_3 =0$$ and
$$g =4(e_1 e_2 + e_2 e_3 + e_1 e_3)$$

Prove that $$
g^2=16(e_1^2e_2 ^2 + e_2^2 e_3^2+e_1^2 e_3^2)
$$

What I got :
$$
g^2 =16(e_1^2e_2 ^2 + e_2^2 e_3^2+e_1^2 e_3^2+2e_1^2e_2e_3+2e_1e_2^2e_3+2e_1e_2e_3^2)
$$
However, I fail to see how $2e_1^2e_2e_3+2e_1e_2^2e_3+2e_1e_2e_3^2 = 0$
Would appreciate any help

REPLY [10 votes]: You can factor out $2e_{1}e_{2}e_{3}$ from $$2e_{1}^{2}e_{2}e_{3} + 2e_{1}e_{2}^{2}e_{3} + e_{1}e_{2}e_{3}^{2},$$ which gives you $$2e_{1}e_{2}e_{3}(e_{1} + e_{2} + e_{3}).$$ Because $e_{1} + e_{2} + e_{3} = 0$, then $2e_{1}^{2}e_{2}e_{3} + 2e_{1}e_{2}^{2}e_{3} + e_{1}e_{2}e_{3}^{2} = 0$.

REPLY [9 votes]: Hint :
$$2e_1^2e_2e_3+2e_1e_2^2e_3+2e_1e_2e_3^2 = 2e_1e_2e_3(e_1+e_2+e_3)$$