Question

If the cube root of unity are $1,\omega ,{\omega }^2$ then the root of the equation $(x-1{)}^3-8=0$.

Answer 1

Given the equation is

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

or, $(x-1{)}^3=2^3$.......(1).

As we are given the root of the equation $x^3=1$ are $1,\omega ,{\omega }^2$.

Then the roots of equation (1) are $2+1,2\omega +1,2{\omega }^2+1$.