$x_1$ , $x_2$ 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) ?
 A: Using that :

*

*$x^2=-x+5$


*$x^3=x\cdot x^2=-x^2+5x=x-5+5x=6x-5$
Then:
$$
\require{cancel}
y = \frac{1}{(x+1)^3}=\frac{1}{x^3+3x^2+3x+1}=\frac{1}{(6x-5)+3(-x+5)+3x+1} = \frac{1}{6x+11}
\\ \iff\;\;\;\; 6x = \frac{1}{y}-11
$$
Substituting the latter in $36x^2 + 36x - 180 = 0$ and multiplying by $y^2$ gives:
$$
(1-11y)^2 + 6y(1-11y) - 180y^2 = 0 \;\;\;\;\iff\;\;\;\; 125 y^2 + 16 y - 1 = 0
$$
A: 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 $$\bf 125x^2+16x-1=0$$
$\blacksquare$
A: 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:
$$\begin{eqnarray*}& & \left((x_1 + 1)^3 x - 1\right) \left((x_2 + 1)^3 x - 1\right) \\
& = & \left( (6x_1 + 11) x - 1 \right) \left( (6x_2 + 11) x - 1 \right) \\ 
& = & \left(36 x_1 x_2 + 66(x_1 + x_2) + 121\right) x^2 - \left(6 (x_1 + x_2) + 22\right) x + 1 \\
& = & -125 x^2 - 16 x + 1\end{eqnarray*}$$
A: 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 $125x^2+16x=1$
A: 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)} \ \ . $
