Question

If $1,\omega ,{\omega }^2$ be the cube roots of unity find the roots of the equation $(x-1{)}^3+8=0$

Answer 1

$(x-1{)}^3+8=0$

$\Rightarrow (1-x{)}^3=8$

$\Rightarrow {\left(\frac{1-x}{2}\right)}^3=1$

$\Rightarrow \frac{1-x}{2}=1$ or $\omega$ or ${\omega }^2$

$\therefore x=-1$ or $1-2\omega$ or $1-2{\omega }^2$.