Question

If $\alpha$ and $\beta$ are the root of $x^2+x+1=0$, then ${\alpha }^{16}+{\beta }^{16}=?$

• A. $2$
• B. $1$
• C. $-1$
• D. $\frac{1+\sqrt{3i}}{2}$

Answer 1

The correct option is D $\frac{1+\sqrt{3i}}{2}$

$\begin{array}{c}{x}^2+x+1=0\\ {\lambda }^{16}+{\beta }^{16}=2\\ Distriminent=b^{2}4ac\\ =1-4=-3\\ Roots=\frac{-1\pm \sqrt{-3}}{2}\\ \Rightarrow \lambda =\frac{\left(1+\sqrt{-3}\right)}{2}=\frac{\left(1+\sqrt{3}i\right)}{2}-w\\ \Rightarrow \beta =\frac{\sqrt{-3-1}}{2}=\frac{\left(\sqrt{3i-1}\right)}{2}=w\\ where\left(w\right)iscuberootsofunity\\ \therefore 1+w+w^2=0\\ and{w}^3=1\\ \Rightarrow \therefore {\lambda }^{16}+{\beta }^{16}={\left(w^2\right)}^{16}=w^{16}\left(w^2+1\right)\\ \Rightarrow w^{15}w\left(w^2+1\right)=1\left(w^3+w\right)=1+w=1\frac{\sqrt{3i-1}}{2}\\ \therefore \frac{1+\sqrt{3i}}{2}Ans.\end{array}$