Question

$(x^3-1)$ 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

Answer: (B)

$(x-1)(x-\omega )(x-{\omega }^2)$

$x^3-1=0⟹(x-1)(x^2+x+1)$

Now factorize $x^2+x+1$

Roots of this equation are $\frac{-1+\sqrt{1-4}}{2},\frac{-1-\sqrt{1-4}}{2}=\frac{-1+\sqrt{3}i}{2},\frac{-1-\sqrt{3}i}{2}$

Let $\omega =\frac{-1+\sqrt{3}i}{2}$

$\Rightarrow {\omega }^2=\frac{1}{4}(1-3-2\sqrt{3}i)=\frac{-1-\sqrt{3}i}{2}$

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

The factored form is hence $(x-1)(x-\omega )(x-{\omega }^2)$