$x_1$ , $x_2$ 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) ?

I guess I actually "went there" by looking at the roots themselves:  for $ \ x^2 +  x - 5 \ = \ 0 \ \ , $ we have $ \ x_{1,2} \ = \ -\frac12 \ \pm \ \frac{\sqrt{21}}{2} \ \ . $  For the equation choices, all of them are of the form $ \ 125·x^2 \ \pm \ \begin{array}{c} 12 \\ 16 \end{array}·x \ - \ 1 \ = \ 0 \ \ ,  $ so the middle term is chiefly what we are going to want to distinguish.  We will concentrate on the monic polynomial $ \  x^2 \ \pm \   \frac{12? \   16?}{125}  ·x \ - \ \frac{1}{125} \ \ $ and see what the transformed zeroes prove to be.
For numbers of the form  $ \ x_{1,2} \ = \ -\frac12 \ \pm \ \frac{\sqrt{D}}{2} \ \ , $ we find
$$ x_{1,2} + 1 \ \ = \ \ +\frac12 \ \pm \ \frac{\sqrt{D}}{2} \ \ \Rightarrow \ \ \frac{1}{x_{1,2} + 1} \ \ = \ \ \frac{2}{1 \ \pm \ \sqrt{D}} \ \ = \ \   \frac{2·(1 \ \mp \ \sqrt{D})}{1 \ - \ D} $$
$$ \Rightarrow \ \ x'_{1,2} \ \ = \ \ \left( \frac{1}{x_{1,2} + 1} \right)^3 \ \ =  \ \   \frac{8·(1 \ \mp \ \sqrt{D})^3}{(1 \ - \ D)^3} \ \ . $$
We pretty definitely don't want to have to multiply these expressions out, but that won't be necessary.  The middle coefficient of the new monic polynomial is $ \ -(x'_1 \ + \ x'_2) \ = \ \frac{8}{(1 \ - \ D)^3} \ · \ [ \ (1 \ + \ \sqrt{D})^3 \ + \ (1 \ - \ \sqrt{D})^3 \ ] \ \ . $  We don't even really need to carry out the binomial expansions of the terms in brackets, since it is clear that the terms with odd powers of $ \ \sqrt{D} \ $ will "cancel" in the sum (we do need to know the binomial coefficients). The middle coefficient of the monic polynomial is therefore
$$   -(x'_1 \ + \ x'_2) \ \ = \ \ \frac{8}{(1 \ - \ D)^3} \ · \ 2 · (1 \ + \ 3D )  \ \ ; $$
with $  \ D \ =  \ 21 \ \ $ for our problem, we find
$$   -(x'_1 \ + \ x'_2) \ \ = \ \ \frac{8 \ · \ 2 \ · \ (1 \ + \ 3·21 )}{(1 \ - \ 21)^3}  \ \ = \ \ \frac{16 \ · \ 64}{ - 20^3} \ \ = \ \ -\frac{4^2 \ · \ 4^3}{ 4^3 · 5^3} \ \ = \ \ -\frac{16}{125} \ \ . $$
[With just a bit more -- but unrequired -- effort, we can confirm that
$$    x'_1 \ · \ x'_2  \ \ = \ \ \frac{8^2 \ · \ [ \ (1 \ + \ \sqrt{D})  \ · \ (1 \ + \ \sqrt{D}) \ ]^3}{(1 \ - \ D)^6}  \ \   = \ \ \frac{4^3 \ · \  \ (1 \ - \ D)^3}{(1 \ - \ D)^6}\ \  = \ \ \frac{4^3 }{(1 \ - \ D)^3} $$ $$ = \ \ \frac{ 4^3}{ - 4^3 · 5^3} \ \ = \ \ -\frac{1}{125} \ \ . \ ] $$
Hence, the transformed quadratic equation is $ \ 125·\left(x^2 \ + \ \frac{16}{125}·x \ - \ \frac{1}{125} \right) \ = \ 0 \ \ , $ which is choice $ \ \mathbf{(1)} \ \ . $
