# Showing an equality of complex numbers

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

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 $$2e_{1}^{2}e_{2}e_{3} + 2e_{1}e_{2}^{2}e_{3} + e_{1}e_{2}e_{3}^{2} = 0$$.
Hint : $$2e_1^2e_2e_3+2e_1e_2^2e_3+2e_1e_2e_3^2 = 2e_1e_2e_3(e_1+e_2+e_3)$$