Let $\omega \ne 1,$ be a cube root of unity, and $a,b\in R$.

Statement$-1$: $a^3+b^3=(a+b)\left(a\omega +b{\omega }^2\right)\left(a{\omega }^2+b\omega \right)$

Statement$-2$: $x^3-1=(x-1)\left(x{\omega }^2-\omega \right)\left(x\omega -{\omega }^2\right)$ for each $x\in R.$

We have $–x$
$\begin{array}{l}\left(x{\omega }^2-\omega \right)\left(x\omega -{\omega }^2\right)\\ =x^2{\omega }^3-x{\omega }^2-x{\omega }^4+{\omega }^3\\ =x^2-\left({\omega }^2+\omega \right)x+1\\ =x^2+x+1\end{array}\left[\because {\omega }^2+\omega =-1\right]$

$\begin{array}{l}\therefore (x-1)\left(x{\omega }^2-\omega \right)\left(x\omega -{\omega }^2\right)\\ =(x-1)\left(x^2+x+1\right)=x^3-1\end{array}$

Thus, Statement$-2$is true. Replacing $x$ by –x, we get

$\begin{array}{l}(-x{)}^3-1=(-x-1)\left(-x{\omega }^2-\omega \right)\left(-x\omega -{\omega }^2\right)\\ \Rightarrow x^3+1=(x+1)\left(x{\omega }^2+\omega \right)\left(x\omega +{\omega }^2\right)\end{array}$

Taking conjugate of both the sides, we get

        $x^3+1=(x+1)\left(x\omega +{\omega }^2\right)\left(x{\omega }^2+\omega \right)$
                                                           $[\therefore \omega ={\omega }^2]$
If $b=0,$ , statement$-1$ is clearly true. Suppose $b\ne 0.$
Replacing $x$ by $a/b$ we get 

     $\begin{array}{r}{\left(\frac{a}{b}\right)}^3+1=\left(\frac{a}{b}+1\right)\left(\frac{a}{b}w+w^2\right)\left(\frac{a}{b}{w}^2+w\right)\\ \Rightarrow a^3+b^3=(a+b)\left(aw+b{w}^2\right)\left(a{w}^2+bw\right)\end{array}$

Thus, statement$-1$ is also true and statement$-2$ is a correct explanation for it.