SOLUTION
We want to write
[tex]\begin{gathered} \mleft(3-2i\mright)^3\text{ in simplest form } \\ a+bi \end{gathered}[/tex]This means we have to expand
[tex](3-2i)^3[/tex]Applying perfect cube formula, we have
[tex]\begin{gathered} \mleft(a-b\mright)^3=a^3-3a^2b+3ab^2-b^3 \\ \text{where } \\ a=3,\: \: b=2i \end{gathered}[/tex]We have
[tex]\begin{gathered} (a-b)^3=a^3-3a^2b+3ab^2-b^3 \\ \mleft(3-2i\mright)^3=3^3-(3\times3^2\times2i)+(3\times3\times(2i)^2)-(2i)^3_{} \\ =27-(27\times2i)+(9\times(2i)^2)-(2i)^3_{} \end{gathered}[/tex]This becomes
[tex]\begin{gathered} \text{note that i = }\sqrt[]{-1} \\ i^2=\sqrt[]{-1^2}=-1 \\ So\text{ we have } \\ =27-(27\times2i)+(9\times(2i)^2)-(2i)^3_{} \\ 27-54i+(9\times4i^2)-(8i^2\times i) \\ 27-54i+(9\times4\times-1)-(8\times-1\times i) \\ 27-54i-36+8i \\ -9-46i \end{gathered}[/tex]Hence the answer is
[tex]-9-46i[/tex]