Question

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

• 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

The correct option is B

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

Explanation for the correct option:

Step 1: Simplify the equation

Given that, $1,\omega ,{\omega }^2$ are the cube root of unity.

Solve the equation ${\left(x-1\right)}^3+8=0$ to find its roots.

$\begin{array}{rcl}\therefore \left(x-1\right){}^3+8& =& 0\\ & \Rightarrow & {\left(x-1\right)}^3=-8\end{array}$

Take cube root both sides we get:

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

Step 2: Find the roots

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

The first root will be,

$\begin{array}{rcl}\therefore x-1& =& -2\\ & \Rightarrow & x=-1\end{array}$

The second root will be,

$\begin{array}{rcl}\therefore x-1& =& -2\omega \\ & \Rightarrow & x=1-2\omega \end{array}$

The third root will be,

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

Hence, option B is correct.