Question

.$\left(x^3-1\right)$ can be factorised as: where $\omega$ is one of the cube roots of unity.

• A. $(x-1)(x–\omega )(x+{\omega }^2)$
• B. $(x-1)(x–\omega )(x-{\omega }^2)$
• C. $(x-1)(x+\omega )(x+{\omega }^2)$
• D. $(x-1)(x+\omega )(x-{\omega }^2)$

Answer 1

The correct option is B

$(x-1)(x–\omega )(x-{\omega }^2)$

Explanation for the correct option.

We know that

.$\left(x^3-1\right)=\left(x-1\right)\left(x^2+x+1\right)$

Now, roots of $x^2+x+1$ are:

$$\begin{gathered} x=\frac{-1\pm \sqrt{1^2-4\times 1\times 1}}{2\times 1} \\ =\frac{-1\pm \sqrt{1-4}}{2} \\ =\frac{-1\pm \sqrt{-3}}{2} \\ =\frac{-1\pm \sqrt{3}i}{2}\left[\because \sqrt{-1}=i\right] \end{gathered}$$

Therefore, $\frac{-1+\sqrt{3}i}{2},\frac{-1-\sqrt{3}i}{2}$ are the roots of $x^2+x+1$.

Now,

$$\begin{gathered} \omega =\frac{-1+i\sqrt{3}}{2} \\ {\omega }^2=\frac{-1-i\sqrt{3}}{2} \end{gathered}$$

Hence $1,\omega ,{\omega }^2$ are the roots.

Thus, factors are $(x–1)(x–\omega )$ and $\left(x-{\omega }^2\right)$.

Hence, the correct option is $\mathrm{B}$.