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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

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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 $2e_{1}^{2}e_{2}e_{3} + 2e_{1}e_{2}^{2}e_{3} + e_{1}e_{2}e_{3}^{2} = 0$.

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Hint : $$2e_1^2e_2e_3+2e_1e_2^2e_3+2e_1e_2e_3^2 = 2e_1e_2e_3(e_1+e_2+e_3)$$

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