Question

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

• A. $-1,1+2\omega ,1+2{\omega }^2$
• B. $-1,1-2\omega ,1-2{\omega }^2$
• C. $-1,-1,-1$
• D. $2,2,2$

Answer 1

Answer: (B)

$-1,1-2\omega ,1-2{\omega }^2$

Let $x-1=\alpha$

$\Rightarrow {\alpha }^3+8=0$

$\alpha =-2$ or $\alpha =-2\omega$ or $\alpha =-2{\omega }^2$.............As 1, $\omega ,{\omega }^2$ are cube roots of unity

Hence,

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