Question

Show that ${(-1+\sqrt{3}i)}^3$ is real number.

Answer 1

According to question,

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

$=-1-3\sqrt{3}i+9+3\sqrt{3}i$ .....As, $(i^3=-i$ & $i^2=-1)$

$=8$,

which is a real number,

Hence proved.