Question

If $\alpha$ is nonreal and $\alpha =\sqrt[5]{51}$ then the value of $2^{|1+\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^{-1}|}$ is equal to

• A. $4$
• B. $2$
• C. $1$
• D. None of these

Answer 1

The correct option is C $1$

We have $1+\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4=0$

$1+\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^{-1}$

$=-{\alpha }^4+\frac{1}{\alpha }$

$=\frac{1-{\alpha }^5}{\alpha }$

$=0$

Hence,

$2^{|1+\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^{-1}|}=1$