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Solve the equation,

$x^3+3x+2i=0$

Where, $i=\sqrt {-1}$

$x=-i$ satisfies the given equation,

$\therefore x^2(x+i)-xi(x+i)+2(x+i)=0$

$\Rightarrow (x+i)(x^2-xi+2)=0$

So, $x=-i$ is one solution. Now consider,

$x^2-xi+2=0$

$x=-i$ again satisfies the above equation,

$\therefore x(x+i)-2i(x+i)=0$

$\Rightarrow (x+i)(x-2i)=0$

So, $x=-i$, $x=2i$ are two more solutions.

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