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If there exists a cubic equation $x^3 + 2x^2 +3x + 1 = 0$ has the roots $a,b,c$ And it is given that

  • $\frac{1}{a^3} + \frac{1}{b^3} - \frac{1}{c^3}$

  • $\frac{1}{a^3} + \frac{1}{c^3} - \frac{1}{b^3}$

  • $\frac{1}{c^3} + \frac{1}{b^3} - \frac{1}{a^3}$

Are the roots of another cubic

$px^3 + qx^2 + rx + s$ And then what will the value of $p+q-12r+s$ be equal to?

my work so far I initially tried establishing the relations of the sum of the roots taken 1,2 and 3 at a time and the coefficients that exists for any polynomial, I did this for the first and second equation but as I thought not much was simplified in the second equation which left me wondering if there was another shorter way, perhaps a trick or an observation that I am not able to make. I also tried expressing the cubic relation of $a,b, c$ in different ways by using the formulas for $(a+b+c)^3$ but that didn't result in much. I don't think I missed anything in the mentioned attempts that could be of significance but do feel free to check.

the answer, p+q-12r+s=5

I wasn't even able to get close to an answer so.

All help is greatly appreciated

Saakshya Devat
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  • The value of your expression can change if you multiply the whole cubic by a constant, and it will still have the same roots. Is that cubic supposed to be monic? (i.e., $p=1$?) – Meow Mix Apr 13 '17 at 02:57
  • No it isn't specified if it is a monic or not. But I guess that's a condition we can impose, can't we? – Saakshya Devat Apr 13 '17 at 03:08

1 Answers1

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I have to sleep, so this is all I have right now. Perhaps I'll update tomorrow? It's too long for a comment anyways.

So, if we assume it's monic, obviously $p$ will be 1. Now, by Vieta's formulas, $-s$ will be the sum of the roots. And the sum of the roots turns out to be:

$\frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3}$

We want to express this in elementary symmetric polynomials, so I used Mathematica's SymmetricReduction function and got that it is equivalent to:

$$-\frac{3 a^2 b^2 c^2+(a b+a c+b c)^3-3 a b c (a+b+c) (a b+a c+b c)}{(abc)^3}$$

Use Vieta's formulas to get this equal to 12. So $s = 12$.

Meow Mix
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