Question

If $\omega$ is a complex cube root of unity, then $225+(3\omega +8{\omega }^2{)}^2+(3{\omega }^2+8\omega {)}^2=$

Answer 1

$225+{(3\omega +8{\omega }^2)}^2+{(3{\omega }^2+8\omega )}^2$

$⟹$ We know,

${(a+b)}^2=a^2+2ab+b^2$

Also,

$\omega ,{\omega }^2$ are cube roots of unity.

$⟹1+\omega +{\omega }^2=0$

${\omega }^3=1$

$⟹225+{\left(3\omega \right)}^2+\left(64{\omega }^4\right)+48{\omega }^3+9{\omega }^4+64{\omega }^2+48{\omega }^3$

$⟹225+{\omega }^2(9+64)+{\omega }^3(48+48)+{\omega }^4(64+9)$

$⟹225+{\omega }^2\left(73\right)+{\omega }^3\left(96\right)+\omega .{\omega }^3\left(73\right)$

$⟹225+{\omega }^2\left(73\right)+1\left(96\right)+\omega .1\left(73\right)$

$⟹225+73(\omega +{\omega }^2)+96$

$⟹225+73(-1)+96$

$⟹$ $248$