Question

If W is a complex cube root of Unity, then show that: (1 – W + W2)

5 + (1 + W – W2)

5 = 32​

Answer 1

Answer:

See the explanation

Step-by-step explanation:

Concept : If $\omega$ is a complex cube root of unity, then

                      $1+\omega+\omega^2=0$ and $\omega^3=1$

Here

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

              $=[(1+\omega^2)-\omega]^5+[(1+\omega)-\omega^2]^5$

              $=[(-\omega)-\omega]^5+[(-\omega^2)-\omega^2]^5$ using $1+\omega+\omega^2=0$

              $=(-2\omega)^5+(-2\omega^2)^5\\=(-2)^5(\omega^5)-(2)^5(\omega^{10})\\=-32(\omega^3\times \omega^2)-32(\omega^9\times \omega)$

             $=-32(1\times \omega^2)-32(1\times \omega)$, using $\omega^3=1$

             $=-32(\omega^2+\omega)\\=-32(-1)=32=\text{RHS}$, since $1+\omega+\omega^2=0$

Hence proved