Question 13 Marks
If $\omega$ is a complex cube root of unity, prove that $\left(1-\omega+\omega^2\right)^6+\left(1+\omega-\omega^2\right)^6=128$.
Answer
View full question & answer→ω is the complex cube root of unity.
$\therefore \omega^3=1 \text { and } 1+\omega+\omega^2=0$
$\text { Also, } 1+\omega^2=-\omega, 1+\omega=-\omega^2$
$\therefore \text { L.H.S. }=\left(1-\omega+\omega^2\right)^6+\left(1+\omega-\omega^2\right)^6$
$=\left[\left(1+\omega^2\right)-\omega\right]^6+\left[(1+\omega)-\omega^2\right]^6$
$=(-\omega-\omega))^6+\left(-\omega^2-\omega^2\right) 6$
$=(-2 \omega)^6+\left(-2 \omega^2\right)^6$
$=64 \omega^6+64 \omega^{12}$
$=64\left(\omega^3\right)^2+64\left(\omega^3\right)^4$
$=64(1)^2+64(1)^4$
$=128$
$=\text { R.H.S. }$
$\therefore \omega^3=1 \text { and } 1+\omega+\omega^2=0$
$\text { Also, } 1+\omega^2=-\omega, 1+\omega=-\omega^2$
$\therefore \text { L.H.S. }=\left(1-\omega+\omega^2\right)^6+\left(1+\omega-\omega^2\right)^6$
$=\left[\left(1+\omega^2\right)-\omega\right]^6+\left[(1+\omega)-\omega^2\right]^6$
$=(-\omega-\omega))^6+\left(-\omega^2-\omega^2\right) 6$
$=(-2 \omega)^6+\left(-2 \omega^2\right)^6$
$=64 \omega^6+64 \omega^{12}$
$=64\left(\omega^3\right)^2+64\left(\omega^3\right)^4$
$=64(1)^2+64(1)^4$
$=128$
$=\text { R.H.S. }$