<s>[INST] Showing $\left(\frac{-1+\sqrt{3}i}{2}\right)^3=1$ 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? 
 [/INST] 
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
$$\left(\dfrac{-1+ i\sqrt{3}}{2}\right)^3=1$$
</s>