Question

If $r$ is non-real and $r=\sqrt[5]{51}$, then the value of $2^{|1+r+r^2+r^{-2}-r^{-1}|}$ is equal to

• A. $2$
• B. $4$
• C. $8$
• D. None of these

Answer 1

The correct option is B $4$

$|1+r+r^2+r^{-2}-r^{-1}|=|1+r+r^2+r^3-r^4|$

$[\because r^5=1\Rightarrow r^3.r^2=1\Rightarrow r^{-2}=r^3$ and $r^4.r=1\Rightarrow r^{-1}=r^4]$

$=|1+r+r^2+r^3+r^4-2r^4|=|\frac{1-r^5}{1-r}-2r^4|=|0-2r^4|[\because r^5=1]$

$=2{\left|r\right|}^4=2\left(1\right)=2[\because \left|r\right|=1$ as $r^5=1]$

$\therefore 2^{|1+r+r^2+r^{-2}-r^{-1}|}=2^2=4$