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

Answer 1

Answer: (B)

Since ${\left(x-1\right)}^3+8=0$
$\Rightarrow {\left(x-1\right)}^3=-8={\left(-2\right)}^3$
$\Rightarrow {\left(\frac{x-1}{-2}\right)}^3=1$
$\Rightarrow \left(\frac{x-1}{-2}\right)={\left(1\right)}^{1/3}$
$\therefore$ roots of $\left(\frac{x-1}{-2}\right)$ are $1,\omega$ and ${\omega }^2.$
$\Rightarrow$ roots of $\left(x-1\right)are-2,-2\omega$ and $-2{\omega }^2$
$\Rightarrow$ roots of $x$ are $-1,1-2\omega$ and $1-2{\omega }^2$
Note : If $1,\omega$ and ${\omega }^2$ are cube roots of unity, then $1+\omega +{\omega }^2=0$ and ${\omega }^3=1$