Question

If $x^2-x+1=0$, then the value of ${(x+\frac{1}{x})}^2+{(x^2+\frac{1}{x^2})}^2+{(x^3+\frac{1}{x^3})}^2+{(x^5+\frac{1}{x^5})}^2$ is equal to

Answer 1

$x^2-x+1=0\Rightarrow x+\frac{1}{x}=1$
Squaring both sides, we get $x^2+\frac{1}{x^2}=-1$
Cubing, we get $x^3+\frac{1}{x^3}+3(x+\frac{1}{x})=1$
or $x^3+\frac{1}{x^3}=-2$
Also, $(x^2+\frac{1}{x^2})(x^3+\frac{1}{x^3})=x^5+\frac{1}{x^5}+x+\frac{1}{x}$
$\Rightarrow (-1)(-2)=x^5+\frac{1}{x^5}+1$ or $x^5+\frac{1}{x^5}=1$
$\therefore {(x+\frac{1}{x})}^2+{(x^2+\frac{1}{x^2})}^2+{(x^3+\frac{1}{x^3})}^2+{(x^5+\frac{1}{x^5})}^2$
$=1+1+4+1=7$