Question

if n is an even number and \[\large (x+\frac{1}{x}+1)(x+\frac{1}{x})=1-x^3-x^4-...-x^{n-3}\]....find \[\large (x^n+\frac{1}{x^n}+1)(\frac{1}{x^n}+x^n)\]

Answer 1

let x^n = a

Answer 2

just a hunch ....

Answer 3

\[(x+\frac{1}{x}+1)(x+\frac{1}{x})=x^2+\frac{1}{x^2}+2+x+\frac{1}{x}\]
if we multiply by x^2 we get\[\large x+2x^2+x^3+x^4=x^2(1-x^3-x^4...-x^{n-3})=x^2-x^5-x^6...-x^{n-1}\]
by 1 we get
\[0=x+x^2+x^3+x^4+x^5+x^6+...+x^{n-1}=x\frac{x^{n}-1}{x-1}\]
\[\fbox{1 }\\a+ar+r^2+...+ar^{n-1}=s_n=a\frac{r^n-1}{r-1}\]

since
\[x\frac{x^{n}-1}{x-1}=0\implies x\ne0,x\ne1\\x(x^n-1)=0\implies x^n=1\]
so\[\color{blue}{\large (x^n+\frac{1}{x^n}+1)(\frac{1}{x^n}+x^n)=(1+\frac{1}{1}+1)(1+\frac{1}{1})}=3*2=6\]