$S_{1}\iff S_{2}$ in complex numbers

Question

Let : $a_0 , a_1 , a_2 , b_0 , b_1 , b_2 \in \mathbb{C} $ :

Show the following equivalence :

$$\begin{cases} ( 1 + a_0 ) ( 1 + a_1 ) ( 1 + a_2 ) &=& ( 1 + b_0 ) ( 1 + j b_0 ) ( 1 + j^2 b_0 ) \\ ( 1 + j a_0 ) ( 1 + j a_1 ) ( 1 + j a_2 ) &=& ( 1 + b_1 ) ( 1 + j b_1 ) ( 1 + j^2 b_1 ) \\ ( 1 + j^2 a_0 ) ( 1 + j^2 a_1 ) ( 1 + j^2 a_2 ) &=& ( 1 + b_2 ) ( 1 + j b_2 ) ( 1 + j^2 b_2 ) \end{cases} $$

$$ \ \ \Longleftrightarrow \ \ $$

$$ \begin{cases} ( 1 + b_0 ) ( 1 + b_1 ) ( 1 + b_2 ) &=& ( 1 + a_0 ) ( 1 + j a_0 ) ( 1 + j^2 a_0 ) \\ ( 1 + j b_0 ) ( 1 + j b_1 ) ( 1 + j b_2 ) &=& ( 1 + a_1 ) ( 1 + j a_1 ) ( 1 + j^2 a_1 ) \\ ( 1 + j^2 b_0 ) ( 1 + j^2 b_1 ) ( 1 + j^2 b_2 ) &=& ( 1 + a_2 ) ( 1 + j a_2 ) ( 1 + j^2 a_2 ) \end{cases}$$

with : $j = e^{\frac{2i \pi}{3}}$.

any help would be appreciated

Answer 1

The result is false. Let $a_0 = 1, a_1 = i, a_2 = -i, b_0 = \sqrt[3]{3}, b_1 = 0, b_2 = 0$.