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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...

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1 Answer 1

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Let the equation $$6y^3 - 7y^2 + 3y - 1 = 0\tag{*}$$ have roots $\alpha$, $\beta$ and $\gamma$.

Using Vieta's formulas: $$\sum \alpha = \alpha + \beta + \gamma = - \frac{-7}{6} = \frac{7}{6}\tag{1}$$ $$\sum \alpha\beta = \alpha\beta + \beta\gamma + \gamma\alpha = \frac{3}{6} = \frac{1}{2}$$

Using the identity $\sum \alpha^2 = \left ( \sum\alpha \right )^2 - 2\sum\alpha\beta$ gives $$\sum \alpha^2 = \frac{49}{36} - 2\times\frac{1}{2} = \frac{13}{36}\tag{2}$$

Rearranging $(*)$ $$6y^3 = 7y^2 - 3y + 1$$ so that $$6\left (\sum\alpha^3\right ) = 7\left(\sum\alpha^2\right) - 3\left (\sum\alpha\right ) + 1 + 1 + 1.$$ (after adding $6\alpha^3 = 7\alpha^2 - 3\alpha + 1$, $6\beta^3 = 7\beta^2 - 3\beta + 1$ and $6\gamma^3 = 7\gamma^2 - 3\gamma + 1$)

Applying $(1)$ and $(2)$ $$\sum\alpha^3 = \frac{1}{6}\left ( 7\times\frac{13}{36} - 3\times\frac{7}{6} + 3 \right ) = \frac{73}{216}$$

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