Question

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

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

Answer 1

Answer: (D)

$54$

$x^2+x+1=0$ implies

$x=w,w^2$ where $w,w^2$ are cube roots of unity.

Now

$1+w+w^2=0$
$w^3=1$

And

$\overline{w}=w^2$.

Now

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

$=(w+\overline{w}{)}^2+(w^2+\overline{w^2}{)}^2+...(w^{27}+\overline{w^{27}}{)}^2$

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

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

$=36+18(-1{)}^2$

$=36+18$
$=54$