<s>[INST] 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.
 [/INST] 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}$.
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