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) ?
Best Answer
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:
$$ (1-11y)^2 + 6y(1-11y) - 180y^2 = 0 \;\;\;\;\iff\;\;\;\; 125 y^2 + 16 y - 1 = 0 $$