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Given the polynomial $x^{3}-p{x}+1=0$, evaluate $\frac{bc}{a^{5}}+\frac{ac}{b^{5}}+\frac{ab}{c^{5}}$ in terms of $p$ if $a,b$ and $c$ are the roots of the polynomial.

My attempt involved:

$$\frac{bc}{a^{5}}+\frac{ac}{b^{5}}+\frac{ab}{c^{5}}=\frac{\left(ab+ac+bc\right)^{6}-4...}{\left(-1\right)^{5}}$$

but the power $6$ is just ridiculous. There must be a way to evaluate this without having to expand a trinomial to the power of $6$ by hand. Any ideas?

Blue
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user71207
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    But it is interesting that the first term is $ \ \frac{abc}{a^6} \ $ and similarly for the other two. And $ \ x^6 \ = \ (px-1)^2 \ $ ... –  Apr 14 '21 at 05:12
  • The expression is equal to $-(u^6+v^6+w^6)$ where $u, v, w$ are the roots of $y^3-py^2+1=0$. – Martin R Apr 14 '21 at 05:31
  • @boojum hmmm ok that way might be shorter than mine. Ill give it a go – user71207 Apr 14 '21 at 05:39
  • @boojum If I factor out the $abc$ I get $-1(\frac{1}{a^6}+\frac{1}{b^6}+\frac{1}{c^6})$ which is back to square one? I assume that was not your intention then? – user71207 Apr 14 '21 at 10:09
  • To whoever edited my post, what does the $cyc$ in $\sum_{cyc}$ mean? – user71207 Apr 15 '21 at 04:20

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As Martin R noted we have to find the value of $$t_6=p^6+q^6+r^6$$ where $p.q.r$ are the roots of $x^3-px^2+1$. Now let $t_k=p^k+q^k+r^k$ for $k\ge 3$ indeed from $x^3=px^2-1$ we get the recurrence $$t_k=pt_{k-1}-3$$ Now we have to find $t_6$ when we know the value of $t_2=p^2$....

The rest should be easy

Albus Dumbledore
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  • Ah, a recurrence method is brilliant. I don't completely understand the "new" roots you and Martin R have created - is it just a substitution $p = \frac{1}{a^6}$ etc? – user71207 Apr 14 '21 at 10:13
  • @user71207 I replaced $x\to \frac{1}{x}$ in the equation so that the roots became $p=1/a,q=1/b,r=1/c$ – Albus Dumbledore Apr 14 '21 at 10:34
  • Ah yes, that's what a meant - a typo on my part. And then I just resubstitute $\frac{1}{x} \rightarrow x$ after? – user71207 Apr 14 '21 at 10:37
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    Thanks! you always have the best answers for my polynomial questions lol – user71207 Apr 14 '21 at 10:38
  • I just noticed that it should be $x^3 = px -1$, instead of $x^3 = px^2 -1$. How did you get the $3$ in the recurrence formula as well? The new one would be $t_k = pt_{k-2} -1$??? Or minus $3$? – user71207 Apr 27 '21 at 12:03
  • I think I see where the minus $3$ comes from. $t_{k-3}$ which is equal to $p^0 + q^0 + r^0 = 3$ – user71207 Apr 27 '21 at 12:07
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    @user71207 you got that right, note that i replaced x by 1/x and then multiplying by x^3 to get rif of denominator we get $$x^3-px^2+1=0$$ – Albus Dumbledore Apr 27 '21 at 12:09
  • Oh my bad, you were indeed always correct. Forgot about $x \rightarrow \frac{1}{x}$ – user71207 Apr 27 '21 at 12:12
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    @user71207 although i was correct this time that may not be the case allthe time ,all humans make mistakes :) – Albus Dumbledore Apr 27 '21 at 12:14