Write the three cube roots of -27 in rectangular coordinates. Give exact answers.

$$\sqrt[3]{-27}=\sqrt[3]{27}\sqrt[3]{-1}=3\sqrt[3]{-1}$$

$$\sqrt[3]{-1}=\sqrt[3]{e^{\imath \pi}}=\sqrt[3]{e^{\imath (\pi +2k \pi)}}~~k\in \mathbb{Z}$$

and we pick values corresponding to k=0,1,2

$$\sqrt[3]{-1}=e^{\imath \pi/3}$$

$$\sqrt[3]{-1}=e^{\frac{\imath(\pi + 2\pi)}{3}}=e^{\imath\pi}=-1$$

$$\sqrt[3]{-1}=e^{\frac{\imath(\pi + 4\pi)}{3}}=e^{\imath 5\pi/3}=e^{-\imath \pi/3}$$

so

$$\sqrt[3]{-27}=-3, ~3e^{\imath \pi/3}, ~3e^{-\imath \pi/3}$$

Oh I see it wants the answers in rectangular form

To do this, convert the complex exponential into rectangular form as follows

$$e^{\imath x}=\cos(x)+\imath \sin(x)$$

$$-3 = -3 \\$$

$$3e^{\imath \pi/3} = 3(1/2 + \imath \sqrt{3}/2)=3/2+\imath 3\sqrt{3}/3$$

$$3e^{-\imath \pi/3} = 3(1/2 - \imath \sqrt{3}/2)=3/2-\imath 3\sqrt{3}/3$$

.