If $\frac{(1+x)^2}{1+x^2}=\frac{13}{37}$, then find the value of $\frac{(1+x)^3}{1+x^3}$ 
Let $x$ be a real number such that $\frac{(1+x)^2}{1+x^2}=\frac{13}{37}$. What is the value of $\frac{(1+x)^3}{1+x^3}$?

Of course, one way is that to solve for $x$ from the quadratic $37(1+x)^2=13(1+x^2)$ which gives the value $x=\frac{-37\pm\sqrt{793}}{24}$. Thus we can compute $\frac{(1+x)^3}{1+x^3}=\frac{13}{49}$.
But I am trying to find the result from the equation $\frac{(1+x)^2}{1+x^2}=\frac{13}{37}$ without solving for $x$ using some algebra. Here are my attempts to do that:
We have $$\begin{align} & \frac{(1+x)^2}{1+x^2}=\frac{13}{37}\\ &\implies \frac{(1+x)^2}{(1+x)^2-2x}=\frac{13}{37}\\ &\implies \frac{(1+x)^3}{(1+x)^3-2x(1+x)}=\frac{13}{37}\\ &\implies \frac{(1+x)^3}{1+x^3+3x(2+x)-2x(1+x)}=\frac{13}{37}\\ &\implies \frac{(1+x)^3}{1+x^3+x(4+x)}=\frac{13}{37}
\end{align}$$
I can't seem to proceed from here.
 A: Given
$$\frac{(1+x)^2}{1+x^2}=a,$$
we have
$$\frac{(1+x)^3}{1+x^3}=\frac{(1+x)^3}{(1+x)(1-x+x^2)}=\frac{(1+x)^2}{1-x+x^2}.$$
Now,
$$\frac{1-x+x^2}{(1+x)^2}=\frac{\frac32(1+x^2)-\frac12(1+x)^2}{(1+x)^2}=\frac{3}{2a}-\frac12,$$
so the answer is
$$\frac1{\frac{3}{2a}-\frac12}=\frac{2a}{3-a}.$$
A: Instead of $ \displaystyle \frac{(1+x)^3}{1+x^3}$, I first find $ \displaystyle \frac{1+x^3}{(1+x)^3}$
$ \displaystyle \frac{1+x^3}{(1+x)^3} = \frac{(1+x) (1+x^2 -x)}{(1+x)^3} = 1 - \frac{3x}{(1+x)^2} \tag1$
(as $x \ne -1$)
Now note that $ \displaystyle \frac{1+x^2}{(1+x)^2} = \frac{37}{13} \implies 1 - \frac{2 x}{(1+x)^2} = \frac{37}{13}$
So we know $ \displaystyle \frac{x}{(1+x)^2} = - \frac{12}{13}$ and we can plug into the first equation which gives us $ \displaystyle \frac{49}{13}$. That leads to the answer of $ \displaystyle \frac{13}{49}$.
A: To go a little further, let's calculate all $f(n)=\dfrac{1+x^n}{(1+x)^n}$
I take the inverse of the desired quantity because it makes it easier to have a factorizable expression on denominator.
Notice that $f(1)=1$ therefore:
$f(n)=f(n)f(1)=\dfrac{(1+x^n)(1+x)}{(1+x)^{n+1}}=\dfrac{(1+x^{n+1})+x(1+x^{n-1})}{(1+x)^{n+1}}=f(n+1)+\dfrac x{(1+x)^2}f(n-1)$
Though notice that $\alpha=\dfrac{x}{(1+x)^2}=\dfrac{1-f(2)}{2}=-\frac{12}{13}$
You end up with a linear recurrence with constant coefficients:
$$f(n+1)-f(n)+\alpha f(n-1)=0\qquad f(0)=2\quad f(1)=1\tag{E}$$
Which solves to

$$f(n)=r^n+\bar r^n\qquad r,\bar r=\frac 12\pm\frac 1{26}\sqrt{793}$$

Note: the given roots are solutions of $x^2-x+\alpha=0$
And you are able to effectively calculate $\dfrac{(1+x)^n}{1+x^n}=\dfrac 1{f(n)}$ for any given $n$. Of course for $f(3)$ it is quicker to just report in (E).
Remark: since $|\bar r|<1$ then $\bar r^n\to 0$ quickly, so for large values of $n$ we have $f(n)\approx r^n$.
