This is a question in A level Further Pure mathematics pastpaper Nov 2010.
The roots of the equation $x^3+4x-1=0$ are $a$, $b$ and $c$.
i) Use the substitution $y=1/(1+x)$ to show that the equation $6y^3-7y^2+3y-1=0$ has roots $1/(a+1)$, $1/(b+1)$ and $1/(c+1)$.
ii) For the cases $n=1$ and $n=2$, find the value of $1/(a+1)^n+1/(b+1)^n+1/(c+1)^n$
iii) Deduce the value of $1/(a+1)^3+1/(b+1)^3+1/(c+1)^3$
I know how to obtain the answer of i) and ii) by substituting $x=(1-y)/y$ and $\sqrt{y}=1/(1+x)$.
The equation for $n=2$ is $36y^3-13y^2-5y-1=0$
To find the value of the case $n=3$, I used to use a formula such as $S^3=\left(\sum a\right )^3-3\sum a \sum ab+3abc$
But there's another formula given by the Examiner Report for Teachers: $6S^3=7S^2-3S+3$
It seems to be a nice method to solve this problem but it baffles me for a long time and I still can't find out how to get this...