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I had an argument with my brother this morning about finding p^3+q^3+s^3 where p and q and s are the roots of a cubic. What is this value expressed in terms of the coefficients of the cubic?

(Sorry if this was rushed, but I want to win the argument)

Lucien Jaccon
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    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Feb 09 '24 at 13:15
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    you can look up the solution of the general cubic and compute the sum of cubes yourself. Also, if you really were in an argument, prove that by at least saying what were the two sides of that argument. – coffeemath Feb 09 '24 at 13:21
  • You can find the answer in Wikipedia Newton's identities for power sums. – Somos Feb 09 '24 at 13:23

1 Answers1

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Problem: Let the cubic polynomial be $ax^3 + bx^2 + cx + d$ and $p, q, r$ be it’s roots.

To find: $p^3 + q^3 + r^3$

We know the following identity: $p^3 + q^3 + r^3 - 3pqr = (p+q+r)(p^2 + q^2 + r^2 - pq - qr - pr)$

By Vieta’s relations: $$p + q + r = \frac{-b}{a}\\ pq + qr + pr = \frac{c}{a}\\ pqr = \frac{-d}{a}\\$$

The only thing left to find is $p^2 + q^2 + r^2: $

By using the following identity $(p + q + r)^2 = p^2 + q^2 + r^2 + 2(pq + qr + pr)$ we get: $$\frac{b^2}{a^2} = p^2 + q^2 + r^2 + \frac{2c}{a} \\ \frac{b^2}{a^2} - \frac{2c}{a} = p^2 + q^2 + r^2 \implies \frac{b^2 - 2ac}{a^2}$$

By substituting these in the original identity as stated before: $$p^3 + q^3 + r^3 - \frac{3d}{a} = \Bigg(\frac{-b}{a}\Bigg) \Bigg(\frac{b^2 - 2ac}{a^2} - \frac{c}{a}\Bigg)\\ p^3 + q^3 + r^3 = \Bigg(\frac{-b}{a}\Bigg) \Bigg(\frac{b^2 - 3ac}{a^2}\Bigg) - \frac{3d}{a} \implies \frac{-b^3 - 3a^2d + 3abc }{a^3}$$

Edit: Thank you Ayan. $\tag*{$\blacksquare$}$

wrath
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ShauryaK
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