show that (-1+root3i)³ is a real number.

the given expression is:
$$\begin{gathered} {\left(-1+\sqrt{3}.i\right)}^3 \\ =\left(-1+\sqrt{3}i\right).{\left(-1+\sqrt{3}i\right)}^2 \\ =\left(-1+\sqrt{3}i\right).(1+3i^2-2\sqrt{3}i) \\ =\left(-1+\sqrt{3}i\right)(1-3-2\sqrt{3}i) \\ =\left(-1+\sqrt{3}i\right)(-2-2\sqrt{3}i) \\ =\left(-1+\sqrt{3}i\right)(-2)(1+\sqrt{3}i) \\ =-2\left(-1+\sqrt{3}i\right)\left(1+\sqrt{3}i\right) \\ =-2.\left[{\left(\sqrt{3}i\right)}^2-1\right] [\sin ce (A+B\left)\right(A-B)=A^2-B^2 \\ =-2\left[3i^2-1\right] \\ =-2.(-3-1) \\ =-2*(-4)=8 \end{gathered}$$
which is purely real number.

hope this helps you