Question

If $x^2+x+1=0$, then the numerical value of
${(x+\frac{1}{x})}^2+{(x^2+\frac{1}{x^2})}^2+{(x^3+\frac{1}{x^3})}^2+{(x^4+\frac{1}{x^4})}^2+....+{(x^{27}+\frac{1}{x^{27}})}^2$ is equal to

• A. 54
• B. 36
• C. 27
• D. 18

Answer 1

The correct option is A 54

$\begin{array}{c}{x}^2+x+7=0\\ {\left(x+\frac{1}{x}\right)}^2+{\left(x^2+\frac{1}{x^2}\right)}^2+{\left(x^3+\frac{1}{x^3}\right)}^2+..........+{\left(x^{27}+\frac{1}{x^{27}}\right)}^2=\\ x^2+x+1=0\\ \Rightarrow x=w\\ {\left(w+\frac{1}{w}\right)}^2+{\left(w^2+\frac{1}{w^2}\right)}^2+\left(w^3+\frac{1}{w^3}\right)+...........+{\left(w^{27}+\frac{1}{w^{27}}\right)}^2\\ =\frac{{\left(w^2+1\right)}^2}{w^2}+\frac{{\left(w^4+1\right)}^2}{w^4}+{\left(2\right)}^2+.........+{\left(2\right)}^2\\ =4\times 9+1\times 18\\ =36+18=54\\ Hence,\\ optionAiscorrectanswer.\end{array}$