To answer this question, we can proceed as follows:
[tex]z=-\frac{3}{2}+\frac{3\sqrt[]{3}}{2}i^{}\Rightarrow z^3=(-\frac{3}{2}+\frac{3\sqrt[]{3}}{2}i)^3[/tex][tex](-\frac{3}{2}+\frac{3\sqrt[]{3}i}{2})^3=(\frac{-3+3\sqrt[]{3}i}{2})^3=\frac{(-3+3\sqrt[]{3}i)^3}{2^3}[/tex]
We applied the exponent rule:
[tex](\frac{a}{b})^c=\frac{a^c}{b^c}[/tex]
Then, we have:
[tex]\frac{(-3+3\sqrt[]{3}i)^3}{2^3}=\frac{(-3+3\sqrt[]{3}i)^3}{8}[/tex]
Solving the numerator, we have:
[tex](a+b)^3=a^3+b^3+3ab(a+b)[/tex]
[tex](-3+3\sqrt[]{3}i)^3=(-3)^3+(3\sqrt[]{3}i)^3+3(-3)(3\sqrt[]{3}i)(-3+3\sqrt[]{3}i)[/tex][tex]-27+81\sqrt[]{3}i^3-27\sqrt[]{3}i(-3+3\sqrt[]{3}i)[/tex][tex]-27+81\sqrt[]{3}i^3+81\sqrt[]{3}i-27\cdot3\cdot(\sqrt[]{3})^2\cdot i^2[/tex][tex]-27+81\sqrt[]{3}i^2\cdot i+81\sqrt[]{3}i-81\cdot3\cdot(-1)[/tex][tex]-27+81\sqrt[]{3}(-1)\cdot i+81\sqrt[]{3}i+243[/tex][tex]-27-81\sqrt[]{3}i+81\sqrt[]{3}i+243[/tex][tex]-27+243=216[/tex]
Then, the numerator is equal to 216. The complete expression is:
[tex]=\frac{(-3+3\sqrt[]{3}i)^3}{8}=\frac{216}{8}=27[/tex]
Therefore, we have that:
[tex]z^3=(-\frac{3}{2}+\frac{3\sqrt[]{3}}{2}i)^3=27[/tex]
In summary, therefore, the value for z³ = 27 (option B).
