Question

If $1,\omega ,{\omega }^2$ are the three cube roots of unity, find the value of ${(1-\omega +{\omega }^2)}^5+{(1+\omega -{\omega }^2)}^5$.

• A. $32$
• B. $30$
• C. $35$
• D. none of these

Answer 1

The correct option is A $32$

$L.H.S=(1-\omega +{\omega }^2{)}^5+(1+\omega -{\omega }^2{)}^5$

$=(1+\omega +{\omega }^2-2\omega {)}^5+(1+\omega +{\omega }^2-2{\omega }^2{)}^5$

Now, $1+\omega +{\omega }^2=0$

$L.H.S=(-2\omega {)}^5+(-2{\omega }^2{)}^5$

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

Now, ${\omega }^3=1$ & $({\omega }^2{)}^3=1$ and $({\omega }^2{)}^2=\omega$

$\Rightarrow L.H.S=-8\left(4{\omega }^2\right)-8\left(4\omega \right)$

$=-32(\omega +{\omega }^2)$

$\Rightarrow L.H.S=+32$

Hence, proved