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 $\omega^{3}=1$