<s>[INST] $x_1$ , $x_2$ are roots of $x=5-x^2$. Find the equation with roots $\frac1{(x_1+1)^3}$ and $\frac1{(x_2+1)^3}$. 
Suppose $x_1$ , $x_2$ are the roots of the equation $x=5-x^2$.Then
$\dfrac1{(x_1+1)^3}$ and $\dfrac1{(x_2+1)^3}$ are roots of which
equation?
$1)125x^2+16x=1$
$2)125x^2=16x+1$
$3)125x^2=12x+1$
$4)125x^2+12x=1$

I solved this problem with the following approache,
I've denoted the roots of the original equation by $\alpha$ and $\beta$ rather than $x_1$, $x_2$ ,
$S=\alpha+\beta=-1$ and $P=\alpha\beta=-5$. We find $S'$ and $P'$ of the new equation,
$$P'=\dfrac1{[(\alpha+1)(\beta+1)]^3}=\dfrac1{(S+P+1)^3}=-\frac1{125}$$
$$S'=\dfrac{(\beta+1)^3+(\alpha+1)^3}{[(\alpha+1)(\beta+1)]^3}=-\frac1{125}\times\left((\alpha^3+\beta^3)+3(\alpha^2+\beta^2)+3(\alpha+\beta)+2\right)$$
$$=-\dfrac{(S^3-3PS)+3(S^2-2P)+3S+2}{125}=-\dfrac{-16+33-3+2}{125}=-\frac{16}{125}$$
Hence the new equation is $125x^2+16x-1=0$. And the answer is first choice.
This was a problem from a timed exam. So can you solve it with other approaches (preferably quicker one) ?
 [/INST] I don't know if this is faster for you, but here's how I did it.
The first thing you need to notice is that higher powers of roots of the equation can be reduced to linear terms only. If $x_i$ is a root of $x = 5 - x^2$, then:
$$\begin{align*}x_i^2 & = 5 - x_i \\ x_i^3 & = 5x_i - x_i^2 \\ & = 6x_i - 5\end{align*}$$
So we can simplify:
$$\begin{eqnarray*}(x_i + 1)^3 & = & x_i^2 + 3 x_i^2 + 3 x_i + 1 \\
& = & (6x_i - 5) + 3 ( 5 - x_i ) + 3 x_i + 1 \\
& = & 6x_i + 11
\end{eqnarray*}$$
And therefore:
$$\begin{eqnarray*}& & \left((x_1 + 1)^3 x - 1\right) \left((x_2 + 1)^3 x - 1\right) \\
& = & \left( (6x_1 + 11) x - 1 \right) \left( (6x_2 + 11) x - 1 \right) \\ 
& = & \left(36 x_1 x_2 + 66(x_1 + x_2) + 121\right) x^2 - \left(6 (x_1 + x_2) + 22\right) x + 1 \\
& = & -125 x^2 - 16 x + 1\end{eqnarray*}$$
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