to simplify the complex number expression (-(1)/(2) + sqrt(3)2i)^3.

You're on a roll — The problem asks to simplify the complex number expression $\left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right)^3$. This expression is a complex cube root of unity, often denoted as $\omega$. We know that $\omega^3 = 1$. Let's verify this using the binomial expansion formula $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$. Let $a = -\frac{1}{2}$ and $b = \frac{\sqrt{3}}{2}i$. Step 1: Apply the binomial expansion formula. $$ \left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right)^3 = \left(-\frac{1}{2}\right)^3 + 3\left(-\frac{1}{2}\right)^2\left(\frac{\sqrt{3}}{2}i\right) + 3\left(-\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}i\right)^2 + \left(\frac{\sqrt{3}}{2}i\right)^3 $$ Step 2: Calculate each term. First term: $$ \left(-\frac{1}{2}\right)^3 = -\frac{1}{8} $$ Second term: $$ 3\left(-\frac{1}{2}\right)^2\left(\frac{\sqrt{3}}{2}i\right) = 3\left(\frac{1}{4}\right)\left(\frac{\sqrt{3}}{2}i\right) = \frac{3\sqrt{3}}{8}i $$ Third term: First, calculate $\left(\frac{\sqrt{3}}{2}i\right)^2$: $$ \left(\frac{\sqrt{3}}{2}i\right)^2 = \frac{(\sqrt{3})^2}{2^2}i^2 = \frac{3}{4}(-1) = -\frac{3}{4} $$ Now, multiply by $3\left(-\frac{1}{2}\right)$: $$ 3\left(-\frac{1}{2}\right)\left(-\frac{3}{4}\right) = -\frac{3}{2} \cdot -\frac{3}{4} = \frac{9}{8} $$ Fourth term: $$ \left(\frac{\sqrt{3}}{2}i\right)^3 = \left(\frac{\sqrt{3}}{2}\right)^3 i^3 = \frac{3\sqrt{3}}{8}(-i) = -\frac{3\sqrt{3}}{8}i $$ Step 3: Combine all the calculated terms. $$ -\frac{1}{8} + \frac{3\sqrt{3}}{8}i + \frac{9}{8} - \frac{3\sqrt{3}}{8}i $$ Step 4: Group the real and imaginary parts. $$ \left(-\frac{1}{8} + \frac{9}{8}\right) + \left(\frac{3\sqrt{3}}{8}i - \frac{3\sqrt{3}}{8}i\right) $$ Step 5: Simplify the real and imaginary parts. $$ \frac{-1+9}{8} + 0i $$ $$ \frac{8}{8} + 0 $$ $$ 1 $$ The simplified expression is $\boxed{1}$. Send me the next one 📸