Question
If the cube roots of unity are $$1,\omega ,{\omega ^2}$$ then the roots of the equation $${\left( {x - 1} \right)^3} + 8 = 0,$$ are
A.
$$ - 1, - 1 + 2\omega , - 1 - 2{\omega ^2}$$
B.
$$ - 1, - 1, - 1$$
C.
$$ - 1,1 - 2\omega , 1 - 2{\omega ^2}$$
D.
$$ - 1,1 + 2\omega ,1 + 2{\omega ^2}$$
Answer :
$$ - 1,1 - 2\omega , 1 - 2{\omega ^2}$$
Solution :
$$\eqalign{
& {\left( {x - 1} \right)^3} + 8 = 0 \cr
& \Rightarrow \,\,\left( {x - 1} \right) = \left( { - 2} \right){\left( 1 \right)^{\frac{1}{3}}} \cr
& \Rightarrow \,\,x - 1 = - 2\,\,{\text{or }} - 2\omega \,\,{\text{or }} - 2{\omega ^2} \cr
& {\text{or }}x = - 1\,\,{\text{or 1}} - 2\omega \,\,{\text{or 1}} - 2{\omega ^2}. \cr} $$