Question

If $x^2+x+1=0$ and $\omega \ne 1$then the value of ${(1+3\omega +3{\omega }^2)}^8+{(3+\omega +3{\omega }^2)}^8+{(3+3\omega +{\omega }^2)}^8$ equals

• A. $3^8$
• B. 0
• C. $-3^8$
• D. None of these

Answer 1

The correct option is B 0

Let $w$ be the cube root of unity.$\therefore w^3=1\&1+w+w^2=0$ where $w=\frac{-1+i\sqrt{3}}{2}\&w^2=\frac{-1-i\sqrt{3}}{2}$
$1+x+x^2=0x=\frac{-1\pm i\sqrt{3}}{2}=w,w^2$
$z={(1+3w+3w^2)}^8+{(3+w+3w^2)}^8+{(3+3w+w^2)}^8\Rightarrow z={(1-3)}^8+{(-3w+w)}^8+{(-3w^2+w^2)}^8\Rightarrow z=2^8(1+w^8+w^{16})\Rightarrow z=2^8(1+w^8+w^{16})\Rightarrow z=2^8(1+w^2+w)\Rightarrow z=0$