Question

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

• A. $-1,-1,-1$
• B. $-1,-1+2w,-1-2w^2$
• C. $-1,1-2w,1-2w^2$
• D. $-1,1-2w,-1-2w^2$

Answer 1

The correct option is C

$-1,1-2w,1-2w^2$

Finding the roots of the equation:

Given, $(x-1{)}^3+8=0$

$\Rightarrow$ ${(x-1)}^3=-8$

$\Rightarrow$ $(x-1)=(-8){}^{1/3}$

$\Rightarrow$ $\left(x-1\right)=8^{1/3}{(-1)}^{1/3}$

$\Rightarrow$ $x-1=2(-1,-w,-w^2)$

$\Rightarrow$ $x=1+2(-1,-w,-w^2)$

$\Rightarrow$ $x=1-2,1-2w,1-2w^2$

$\Rightarrow$ $\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{1}\mathbf{-}\mathbf{2}\mathit{w}\mathbf{,}\mathbf{}\mathbf{1}\mathbf{-}\mathbf{2}{\mathit{w}}^{\mathbf{2}}$

Hence, Option ‘C’ is Correct.