Question

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

Answer 1

${(x+\frac{1}{x})}^2=3x^2+\frac{1}{x^2}+2=3x^4+x^2+1=0t=x^2{t}^2-t+1=0t=\frac{1\pm i\sqrt{3}}{2}t=w;x=\pm \sqrt{w}{x}^6=w^3=1x^{206}+x^{200}+x^{90}+x^{84}+x^{18}+x^{12}+x^6+1x^2{\times x}^{6\times 34}{+x}^2{\times x}^{6\times 33}+x^{15\times 6}+x^{14\times 6}+x^{3\times 6}+x^{6\times 2}{+x}^6+1x^2+x^2+1+1+1+1+1+1w-w+66$