Showing $\left(\frac{-1+\sqrt{3}i}{2}\right)^3=1$

Question

In my textbook they asked me to show that $$\left(\frac{-1+\sqrt{3}i}{2}\right)^3=1$$

but this is not true, I think.

I put down $$\begin{align} \frac12(-1+\sqrt{3}i)(-1+\sqrt{3}i)(-1+\sqrt{3}i)&=\frac12\left[(1-2\sqrt3i-3)(-1+\sqrt3i\right)]\\ &=\frac12\left[-1+2\sqrt3i+3+\sqrt3i+6-3\sqrt3i\right]\\ &=4\\ \end{align}$$

I checked it twice and I got $4$ what am I doing wrong?

Answer 1

Now you know your mistake! But there is an alternative way to show that

Consider this cubic $$x^3-1=0$$ $$(x-1)(x^2+x+1)=0$$ By applying quadratic formula to $x^2+x+1=0$

$$x=\frac{-1- i\sqrt{3}}{2} \quad\text{or}\quad x=\frac{-1+ i\sqrt{3}}{2}$$

Since $x=\dfrac{-1+ i\sqrt{3}}{2}$ is root of $x^3-1=0$ we have

$$\left(\dfrac{-1+ i\sqrt{3}}{2}\right)^3=1$$

Answer 2

$$\left(\frac{-1+\sqrt{3}i}{2}\right)^3=1$$

$$\frac{1}{2^3}(-1+\sqrt{3}i)(-1+\sqrt{3}i)(-1+\sqrt{3}i)=\frac12\left[(1-2\sqrt3i-3)(-1+\sqrt3i\right]$$

$$=\frac18\left[-1+2\sqrt3i+3+\sqrt3i+6-3\sqrt3i\right]$$

Answer 3

HINT: $$\left(\frac{-1+i\sqrt3}{2}\right)=\cos2\pi/3+i\sin2\pi/3=e^{i\frac{2\pi}{3}}$$