Question

If ${(x+\frac{1}{x})}^2=3$ then find the value of $x^{206}+x^{200}+x^{90}+x^{84}+x^{18}+x^{12}+x^6+1$

Answer 1

${(x+\frac{1}{x})}^2=3$
$\therefore x^2+2+\frac{1}{x^2}=3\Rightarrow x^2+\frac{1}{x^2}=1$
$\therefore x^2+1=x^2$
$\therefore x^4-x^2=-1$
$\therefore x^4-x^2+1=0$
Multiply throughout by $(x^2+1)$ we get,
$(x^2+1)(x^4-x^2+1)=0$
$\therefore x^2+1=0$ ------(1)
We have to find value of $x^{206}+x^{200}+x^{90}+x^{84}+x^{18}+x^{12}+x^6+1$
$=x^{200}(x^6+1)+x^{90}(x^6+1)+x^{12}(x^6+1)+1(x^6+1)$
$=(x^6+1)(x^{200}+x^{90}+x^{12}+1)$
But $x^6+1=0$
$x^{206}+x^{200}+x^{90}+x^{84}+x^{18}+x^{12}+x^6+1=0(x^{200}+x^{90}+x^{12}+1)$
$\therefore x^{206}+x^{200}+x^{90}+x^{84}+x^{18}+x^{12}+x^6+1=0$