Question

Let $\omega$ being cube root of unity, then if $x=a+b,y=a\omega +b{\omega }^2,a{\omega }^2+b\omega$. Then show that $x^2+y^2+z^2=6ab$ and $x^3+y^3+z^3=3(a^3+b^3)$.

Answer 1

Given $x=a+b,y=a\omega +b{\omega }^2,z=a{\omega }^2+b\omega$.

Now $x^2+y^2+z^2$

$=(a^2+2ab+b^2)+(a^2{w}^2+2ab{\omega }^3+b^2{\omega }^4)+(a^2{\omega }^4+2ab{\omega }^3+b^2{\omega }^2)$

$=(a^2+2ab+b^2)+(a^2{w}^2+2ab+b^2\omega )+(a^2\omega +2ab+b^2{\omega }^2)$ [Since ${\omega }^3=1$]

$=a^2(1+\omega +{\omega }^2)+6ab+b^2(1+\omega +{\omega }^2)$

$=6ab$.[ $1+\omega +{\omega }^2=0$]

Now

$x^3+y^3+z^3$

$=(a^3+3a^{2}b+3a{b}^2+b^3)+(a^3{\omega }^3+3a{b}^2{\omega }^4+3a{b}^2{\omega }^5+b^3{\omega }^6)+(a^3{\omega }^6+3a^{2}b{\omega }^5+3a{b}^2{\omega }^4+b^3{\omega }^3)$

$=(a^3+3a^{2}b+3a{b}^2+b^3)+(a^3+3a{b}^2\omega +3a{b}^2{\omega }^2+b^3)+(a^3+3a^{2}b{\omega }^2+3a{b}^2\omega +b^3)$ [Since ${\omega }^3=1$]

$=3(a^3+b^3)+3a^{2}b(a+\omega +{\omega }^2)+3a{b}^2(1+\omega +{\omega }^2)$

$=3(a^3+b^3)$. [Since $1+\omega +{\omega }^2=0$]