Answer
$$\left(x^3-x^2-2x\right)+\left(3x^2-3x-6\right)i$$
Work Step by Step
Expanding the equations by multiplying each term in parenthesis by each of the other terms in the other set of parenthesis, we find:
$$\left(x^3-x^2-2x\right)+i\left(-6-3x+3x^2\right)\\ \left(x^3-x^2-2x\right)+\left(3x^2-3x-6\right)i$$
