Question

Prove that : $x^2+xy+y^2=\left(x-y\omega \right)\left(x-y{\omega }^2\right)$

Answer 1

Considering R.H.S

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

$\Rightarrow x^2-xy{\omega }^2-xy\omega +y^2{\omega }^3$

$\Rightarrow x^2-xy({\omega }^2+\omega )+y^2{\omega }^3$

Since,

${\omega }^2+\omega +1=0$ and ${\omega }^3=1$

Therefore,

$\Rightarrow x^2-xy(-1)+y^2\times 1$

$\Rightarrow x^2+xy+y^2$

L.H.S

Hence, proved.