Question

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

• A. $27$
• B. $72$
• C. $45$
• D. $54$

Answer 1

The correct option is D $54$

Given Expression: ${(x+\frac{1}{x})}^2+{(x^2+\frac{1}{x^2})}^2+...+{(x^{27}+\frac{1}{x^{27}})}^2$

$x^2+x+1=0$

As, ${\omega }^2+\omega +1=0$

$\Rightarrow x=worx=w^2$

Also, this equation has a product of roots $w^3=1.$

Taking $x=w$ ....... (1)

$\frac{1}{x}=\frac{1}{w}=\frac{w^2}{w^3}=w^2$ .......... (2)

Substituting equation $\left(1\right)$ and $\left(2\right)$ in the given expression, then we get

$(w+w^2{)}^2+{(w^2+(w^2{)}^2)}^2.....{(w^{27}+(w^2{)}^{27})}^2$

$=(w+w^2{)}^2+(w^2+w^3.w{)}^2.....{\left(w^{27}\right(w^3{)}^9+\left(w^3{)}^{18}\right)}^2$

$=\left(\right(-1{)}^2+(-1{)}^2+2^2)\times 9$...................As ${\omega }^2+\omega =-1$

$=54$