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Let $\alpha,\beta,\gamma$ be three distinct roots of the polynomial $x^3-2x^2-3x-4=0$. Then find $$\frac{\alpha^6-\beta^6}{\alpha-\beta}+\frac{\beta^6-\gamma^6}{\beta-\gamma}+\frac{\gamma^6-\alpha^6}{\gamma-\alpha}.$$ I tried to solve with Vieta's theorem. We have $$\begin{align} \alpha+\beta+\gamma &= 2, \\ \alpha\beta+\beta\gamma+\gamma\alpha &= -3, \\ \alpha\beta\gamma &= 4. \end{align}$$ For example, $\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)=10$ and similarly, we can find $\alpha^3+\beta^3+\gamma^3$...

But it has very long and messy solution. Can anyone help me?

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4 Answers 4

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By long division, the remainder of $x^6$, when divided by $x^3-2x^2-3x-4$, is $77x^2+100x+96$. So we know that $$ \alpha^6=77\alpha^2+100\alpha+96 $$ and the same with other roots. Therefore you are looking at the sum $$ \begin{aligned} S&=\frac{77(\alpha^2-\beta^2)+100(\alpha-\beta)}{\alpha-\beta}+\text{cyclic}\\ &=77(\alpha+\beta)+100+\text{cyclic}\\ &=154(\alpha+\beta+\gamma)+300\\ &=608 \end{aligned} $$ by the (Vieta) relations you have.

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If $\alpha,\beta,\gamma$ satisfy $x^6=ax^2+bx+c$, then the first term is $a(\alpha+\beta)+b$, and we have similar expressions for the others, so the desired sum is $2a(\alpha+\beta+\gamma)+3b=4a+3b$. To find $a$ and $b$, square $x^3$ and reduce modulo the given polynomial.

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  • $\begingroup$ Wow. That's nice idea! Thank you. $\endgroup$
    – Mutse
    May 14, 2020 at 7:30
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Let $\alpha+\beta+\gamma=3u$, $\alpha\beta+\alpha\gamma+\beta\gamma=3v^2$ and $\alpha\beta\gamma=w^3$.

Thus, $$\sum_{cyc}\frac{\alpha^6-\beta^6}{\alpha-\beta}=\sum_{cyc}(2\alpha^5+\alpha^4\beta+\alpha^4\gamma+\alpha^3\beta^2+\alpha^3\gamma^2)=$$ $$=2(243u^5-405u^3v^2+135uv^4+45u^2w^3-15v^2w^3)+$$ $$+81u^3v^2-81uv^4-9u^2w^3+15v^2w^3+$$ $$+27uv^4-18u^2w^3-3v^2w^3=$$ $$=9(54u^5-81u^3v^2+7u^2w^3+24uv^4-2v^2w^3).$$ Now, use your work.

I got $608$.

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  • $\begingroup$ Your solution is nice! But how did you get $2(243u^5-405u^3v^2+135uv^4+45u^2w^3-15v^2w^3)+81u^3v^2-81uv^4-9u^2w^3+15v^2w^3+27uv^4-18u^2w^3-3v^2w^3$ $\endgroup$
    – Mutse
    May 14, 2020 at 7:20
  • $\begingroup$ @Mutse I work with symmetric polynomials and I just know these expressions. But you can get it by the following way: $\alpha^5+\beta^5+\gamma^5=\sum\limits_{cyc}\alpha^3\sum\limits_{cyc}\alpha^2-\sum\limits_{sym}\alpha^3\beta^2=(27u^3-27uv^2+3w^3)(9u^2-6v^2)-(27uv^4-18u^2w^3-3v^2w^3)=...$ $\endgroup$
    – Michael Rozenberg
    May 14, 2020 at 7:25
  • $\begingroup$ @MichaelRozenberg Ok, I got it. Thank you for nice solution! $\endgroup$
    – Mutse
    May 14, 2020 at 7:29
  • $\begingroup$ @Mutse You are welcome! $\endgroup$
    – Michael Rozenberg
    May 14, 2020 at 7:30
  • $\begingroup$ @Mutse I fixed my mistake. Indeed, it's should be equal to $608$. $\endgroup$
    – Michael Rozenberg
    May 14, 2020 at 8:02
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The final answer I got is 608. The answer for this question is provided in the given pdf By Vieta's theorem and using some special algebraic identities that are prevelant in India.. Click on the link .THE whole values are achieved by some easy calculations

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    $\begingroup$ While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. $\endgroup$
    – Calvin Khor
    May 15, 2020 at 5:25

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