Question

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

• A. $-1,-1+2\omega ,-1-2{\omega }^2$
• B. $-1,1-2\omega ,1-2{\omega }^2$
• C. $-1,-1,-1$
• D. None of these

Answer 1

The correct option is B

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

Finding the roots of the equation:

The roots of the equation $(x-1{)}^3+8=0$

$(x-1)=(-2)(1{)}^{\frac{1}{3}}$

Also, $1,\omega ,{\omega }^2$ are the cube roots of unity.

$\begin{array}{rcl}\therefore \mathrm{x}-1& =& -2\text{or}-2{\omega }^2\text{or}-2\omega \\ & \Rightarrow & \mathrm{x}=-1,1-2{\omega }^2,1-2\omega \end{array}$

Hence, option B is correct.