Question

Form the equation whose roots are $m+n,n\omega +n{\omega }^2,m{\omega }^2+n\omega$, where $\omega$ is an imaginary cube root of unity.

Answer 1

$m+n,n\omega +n{\omega }^2,m{\omega }^2+n\omega$

Sum of roots $=m+n+n\omega +n{\omega }^2+m{\omega }^2+n\omega$

$=m(1+\omega +{\omega }^2)+n(1+\omega +{\omega }^2)$

$=m\left(0\right)+n\left(0\right)=0$

Sum of product $=(m+n)(n\omega +n{\omega }^2)+(m{\omega }^2+n\omega )(m{\omega }^2+n\omega )+(m+n)(m{\omega }^2+n\omega )$

$=m^2(1+\omega +{\omega }^2)+n^2(1+\omega +{\omega }^2)+3mn(\omega +{\omega }^2)$

$=m^2\left(0\right)+n\left(0\right)+3mn(-1)=-3mn$

Products of roots $=(m+n)(n\omega +n{\omega }^2)(m{\omega }^2+n\omega )$
$=m^3+n^3+mn(m+n)(1+\omega +{\omega }^2)=m^3+n^3+mn(m+n)=m^3+n^3$

Hence equation is $x^3-0x^2-3mnx-(m^3-n^3)$

or $x^3-3mnx-m^3-n^3$