Question

If $\omega$ and ${\omega }^2$ are complex cube roots of unity, then $1-2{\omega }^2$ is a root of

• A. $x^3-1=0$
• B. $(x-1{)}^3+8=0$
• C. $x^3+1=0$
• D. $(x-1{)}^3-8=0$

Answer 1

The correct option is B $(x-1{)}^3+8=0$

$\begin{array}{c}{\left(x-1\right)}^3+8=0\\ {\left(\frac{x-1}{-2}\right)}^3=1\\ \frac{x-1}{-2}={\omega }^2\\ \Rightarrow x=1-2{\omega }^{2}isrootof{\left(x-1\right)}^3+8=0\\ Hence,optionBisthecorrectanswer.\end{array}$