Question

If $(x+1{)}^2=x,$ then the value of ${(x+\frac{1}{x})}^2+{(x^2+\frac{1}{x^2})}^2+{(x^3+\frac{1}{x^3})}^2+\cdots +{(x^{30}+\frac{1}{x^{30}})}^2$ is

Answer 1

$(x+1{)}^2=x\Rightarrow x^2+x+1=0\Rightarrow x=\frac{-1\pm i\sqrt{3}}{2}=\omega ,{\omega }^2$
where $\omega$ is the cube root of unity.
$\Rightarrow x+\frac{1}{x}=\omega +{\omega }^2=-1$

Now, when $n$ is a multiple of $3$,
$x^n+\frac{1}{x^n}=1+1=2$

When $n$ is not a multiple of $3$,
$x^n+\frac{1}{x^n}=w^n+w^{2n}=-1$
$\therefore {(x+\frac{1}{x})}^2+{(x^2+\frac{1}{x^2})}^2+{(x^3+\frac{1}{x^3})}^2+\cdots +{(x^{30}+\frac{1}{x^{30}})}^2=(-1{)}^2+(-1{)}^2+2^2+(-1{)}^2+(-1{)}^2+2^2+\cdots +(-1{)}^2+(-1{)}^2+2^2$
$=6+6+6+\cdots 10\text{times}=60$