how do i solve (x+2i)^3=2
What about the other solutions?
$\displaystyle (x+2i)^3 = 2$, then all the solutions are:
$\displaystyle x+2i = \sqrt[3]{2},\sqrt[3]{2}\zeta,\sqrt[3]{2}\zeta^2$ where $\displaystyle \zeta$ is primitive root of unity.
That means,
$\displaystyle x = -2i + \sqrt[3]{2} \zeta^k$ for $\displaystyle k=0,1,2$.
I'm searching for them, but I saw that one right when I started trying to solve it.
I'm stuck here $\displaystyle x^{3} + 6x^{2}i - 12x - i - 2 = 0$ Could I use synthetic division somehow to simplify that knowing that $\displaystyle \sqrt[3]{2}-2i$ is a factor? Or is there another way to factor it?
No! Do not fact.
The equation,
$\displaystyle x^3 = a^3$
So,
$\displaystyle x^3 - a^3 = 0$
$\displaystyle (x-a)(x^2+xa+a^2)=0$
So, $\displaystyle x-a=0 \mbox{ and }x^2+xa+a^2 = 0$
So, $\displaystyle x = a, x = \frac{-a\pm \sqrt{a^2 - 4a^2}}{2}$
So in the equation,
$\displaystyle (x+2i)^3 = a^3$ where $\displaystyle a=\sqrt[3]{2}$.
Now use the result above.