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.

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
$$\frac1{\frac{3}{2a}-\frac12}=\frac{2a}{3-a}.$$
