<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] Method-1
Given equation is $x^2+x-5=0$. It's root is $x_1$, hence $x_1^2+x_1-5=0$ $-(1)$
Also note that the equation we have to form has symmetric roots
Therefore think of an expression of type $p(x)=f(x)^2+f(x)-5$ such that $p(\frac{1}{(x_1+1)^3})=x_1^2+x_1+5=0$ which makes $\frac{1}{(x_1+1)^3}$ a root of $p(x)$
Such expression should have $f(\frac{1}{(x_1+1)^3})=x_1$
Clearly $\displaystyle f(x)=\frac{1}{(x)^{\frac{1}{3}}}-1$
Substituting $f(x)$ we get
$p(x)=5(x)^{\frac{2}{3}}+x^{\frac{1}{3}}-1$
And therefore our equation is
$5(x)^{\frac{2}{3}}+x^{\frac{1}{3}}=1$ whch is further simplified to $125x^2+16x=1$
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