# Relations between the roots of a cubic polynomial

How do I solve the last two of these problems?

The roots of the equation $x^3+4x-1=0$ are $\alpha$, $\beta$, and $\gamma$. Use the substitution $y=\dfrac{1}{1+x}$ to show that the equation $6y^3-7y^2+3y-1=0$ has roots $\dfrac{1}{\alpha+1}$, $\dfrac{1}{\beta+1}$, and $\dfrac{1}{\gamma+1}$. $\quad \quad \quad (2)$

For the cases $n=1$ and $n=2$, find the value of $$\dfrac{1}{(\alpha+1)^n}+\dfrac{1}{(\beta+1)^n}+\dfrac{1}{(\gamma+1)^n}. \tag{2}$$ Deduce the value of $\dfrac{1}{(\alpha+1)^3}+\dfrac{1}{(\beta+1)^3}+\dfrac{1}{(\gamma+1)^3}. \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \, \, \,(2)$

Hence show that $\dfrac{(\beta+1)(\gamma+1)}{(\alpha+1)^2}+\dfrac{(\gamma+1)(\alpha+1)}{(\beta+1)^2}+\dfrac{(\alpha+1)(\beta+1)}{(\gamma+1)^2}=\dfrac{73}{36} \quad \quad \quad \quad \quad \quad \quad (3)$

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This is not a single problem. What part looks difficult to you? – Tapu Sep 23 '12 at 15:21
the last part, without solving for $\alpha, \beta, \gamma$ – Monkey D. Luffy Sep 23 '12 at 15:22
How might you relate the expression on the LHS with the previous part? – fretty Sep 23 '12 at 15:37
I've done the rest of the work .. just need help on those parts. I guess that's done too. – Monkey D. Luffy Sep 23 '12 at 15:43

Let $a=\frac1{(1+\alpha)}$ etc,

so, $a,b,c$ are the roots of $6t^3-7t^2+3t-1=0$

$\implies a+b+c=\frac 7 6, ab+bc+ca=\frac 3 6=\frac 12$ and $abc=\frac1 6$

So, $$a^3+b^3+c^3=a^3+b^3+c^3-3abc+3abc$$

$$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc$$

$$=(a+b+c)((a+b+c)^2-3(ab+bc+ca))+3abc$$

$$=(\frac 7 6)((\frac 7 6)^2-3(\frac 12))+3\frac1 6=\frac{73}{216}$$

Now, $$\frac{(\beta+1)(\gamma+1)}{(\alpha+1)^2}=\frac{a^2}{bc}=\frac{a^3}{abc}=6a^3$$

So, $$\sum \frac{(\beta+1)(\gamma+1)}{(\alpha+1)^2}=6(a^3+b^3+c^3)=6\left(\frac{73}{216}\right)=\frac{73}{36}$$

For the generalization of the sums of Powers of Roots, one may look here for the statement, here for the proof.

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$\sum \frac{(\beta+1)(\gamma+1)}{(\alpha+1)^2}$ ?? – Monkey D. Luffy Sep 23 '12 at 15:45
I get the general idea ... perhaps you made typo – Monkey D. Luffy Sep 23 '12 at 15:48
$\frac{(\beta+1)(\gamma+1)}{\alpha+1}=\frac{a^2}{bc}=\frac{a^3}{abc}=6a^3$ this should be like this $\frac{(\beta+1)(\gamma+1)}{(\alpha+1)^2}=\frac{a^2}{bc}=\frac{a^3}{abc}=6a^3$ – Monkey D. Luffy Sep 23 '12 at 15:53
And this $\sum \frac{(\beta+1)(\gamma+1)}{(\alpha+1)^2}$ you copied it wrong!! – Monkey D. Luffy Sep 23 '12 at 15:53
@MonkeyD.Luffy, rectified now. – lab bhattacharjee Sep 23 '12 at 15:55

Part 1,

Since , $\frac{1}{\alpha +1}$ is the root of the equation $6y^3-7y^2+3y-1=0\implies$ $$\frac{1}{(\alpha +1)^3}=\frac{7}{(\alpha +1)^2}-\frac{3}{\alpha +1}+1$$ Similar equality follows for $\beta, \gamma$

Part 2,

Product of roots= $\frac{1}{6}\implies \frac{1}{1+\alpha}\frac{1}{1+\beta}\frac{1}{1+\gamma}=\frac{1}{6}\implies (1+\beta)(1+\gamma)=\frac{6}{1+\alpha}\implies \frac{(1+\beta)(1+\gamma)}{(1+\alpha)^2}=\frac{6}{(1+\alpha)^3}$

Therefore, $\frac{(1+\beta)(1+\gamma)}{(1+\alpha)^2}+\frac{(1+\beta)(1+\gamma)}{(1+\alpha)^2}+\frac{(1+\beta)(1+\gamma)}{(1+\alpha)^2}=6$(result of second part)

Now, you can easily calculate the last sum.

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thanks for part 1 :) ... – Monkey D. Luffy Sep 23 '12 at 15:25
@MonkeyD.Luffy: I have included part 2 also. check it out. – Aang Sep 23 '12 at 16:39

For the first, as Avatar described, you use that $\frac{1}{\alpha+1}$ satisfies the polynomial relation $6y^3-7y^2+3y-1=0$ to express its third power in lower powers of itself.

For the second, you rewrite it simply as $\frac{(\alpha+1)(\beta+1)(\gamma+1)}{(\alpha+1)^3}+\frac{(\alpha+1)(\beta+1)(\gamma+1)}{(\beta+1)^3}+\frac{(\alpha+1)(\beta+1)(\gamma+1)}{(\gamma+1)^3}$ which reduces the problem to calculating $(\alpha+1)(\beta+1)(\gamma+1)$, which being a symmetric rational function in the roots should be easy to relate to the coefficients of $6y^3-7y^2+3y-1$.

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