$x_1$ , $x_2$ 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) ?

As you can see the roots given for the new equation are of the format $\frac{1}{(x+1)^3}$ hence you can use root transformation, either you directly transform $$x \rightarrow \frac{1}{(x+1)^3}$$ or you can do $$x\rightarrow \frac{1}{1+x}$$ and then calculate a bit, I'm going to do using the second method as it reduces the problems and calculations involving irrationalities. Say $$y=\frac{1}{1+x}\implies x=\frac{1-y}{y}$$ Now putting $\bf x$ on the original equation
I get,$$\frac{1-y}{y}=5-\left(\frac{1-y}{y}\right)^2$$ $$\implies5y^2+y-1=0$$
Now, you can observe that the roots of this equation are $\frac{1}{1+x_1},\frac{1}{1+x_2}$ .We see that the roots we want are just these roots raised to the power $3$ Hence, sum of the roots we need is$-(\frac{1}{5})^3+\frac{3}{5}\times\frac {-1}{5}=\frac{-16}{125}$ similarly you can count the product of roots which will be $\frac{-1}{5}$ hence the equation we need is $$\bf 125x^2+16x-1=0$$
$\blacksquare$
