This problem caught my eye in the book yesterday. Till now I still get stuck. Here it is:
If $$\frac{x}{x^2+1}=\frac{1}{3},$$ what is the value of $$\frac{x^3}{x^6+x^5+x^4+x^3+x^2+x+1}?$$
The denominator is a cyclotomic polynomial which can be expressed as $$\frac{x^7-1}{x-1}$$ but I have no idea if this even helps.
The first identity gives: $$ x+\frac{1}{x}=3, $$ hence: $$ x^2+\frac{1}{x^2} = \left(x+\frac{1}{x}\right)^2 - 2 = 7, $$ $$ x^3+\frac{1}{x^3} = \left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)= 18 $$ and $$ x^3+x^2+x+1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3} = 18+7+3+1 = 29 $$ so the answer to your question is $\frac{1}{29}$.
Solve the first equation to get $x = \frac{3 \pm \sqrt{5}}{2}$
Plug in the second to evaluate to $\frac{1}{29}$
It can also solved by this equation $x^2+1=3x$.
$$\frac{x^3}{x^6+x^5+x^4+x^3+x^2+x+1}=\frac{x^3}{3x^5+x^5+x^3+3x+x}= \frac{x^3}{4x^5+4x^3+4x-3x^3}=\frac{x^3}{12x^4+4-3x^3}= \cdots $$
The steps is natural..., I omit; however the answer given by Jack is better.
