Answer
$-2+2i$
Work Step by Step
Use special formula $(a+b)^3=a^3+3a^2b+3ab^2+b^3$.
We have $a=1$ and $b=i$. Thus,
$\begin{align*}
(1+i)^3&=1^3+3\cdot 1^2\cdot i+3\cdot 1\cdot i^2+i^3\\
&=1+3i+3i^2+i^3\end{align*}$
Use $i^3=-i$ and $i^2=-1$ then simplify.
$=1+3i+3(-1)+(-i)$
$=1+3i-3-i$
$=-2+2i$
Hence, the solution in the standard form is $-2+2i$.
