Question

If $\omega$ is a cube root of unity then the value of $(1-\omega +{\omega }^2{)}^4+(1+\omega -{\omega }^2{)}^4$ is:

• A. $-16$
• B. $0$
• C. $-32$
• D. $32$

Answer 1

Answer: (A)

$-16$

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

${(1-\omega +{\omega }^2)}^4+{(1+\omega -{\omega }^2)}^4$

$={(1+{\omega }^2-\omega )}^4+{(1+\omega -{\omega }^2)}^4$

$={(-\omega -\omega )}^4+{(-{\omega }^2-{\omega }^2)}^4$ using the above relations

$={(-2\omega )}^4+{(-2{\omega }^2)}^4$

$=16({\omega }^4+{\omega }^8)$

$=16({\omega }^3\omega +{\omega }^3{\omega }^3{\omega }^2)$

$=16(\omega +{\omega }^2)$ since ${\omega }^3=1$

$=16\omega (1+\omega )$

$=16\omega \left({-\omega }^2\right)$ since $1+\omega =-{\omega }^2$

$=-16{\omega }^3=-16\times 1=-16$