Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

$(x-1{)}^3=-8\Rightarrow x-1=(-8{)}^{1/3}$
$\Rightarrow x-1=-2,-2\omega ,-2{\omega }^2\Rightarrow x=-1,1-2\omega ,1-2{\omega }^2$
$Trick:Byinspection,weseethat\left(b\right)satisfiestheequationi.e,$
$(-1-1{)}^3+8=0(1-2\omega -1{)}^3+8=0(2{\omega }^2-1{)}^3+8=0.$