Find $x_1^3+x_2^3+x_3^3$ for a degree 3 polynomial.

Question

I have the polynomial $P(x) = x^3+mx^2-3x+1, m\in \mathbb{R}$. I need to find $x_1^3+x_2^3+x_3^3$ as a $m$ function.

I tried to use Viette equations: $x_1+x_2+x_3 = m, x_1x_2+x_1x_3+x_2x_3 = 3, x_1x_2x_3 = 1$ Then I expanded $(x_1+x_2+x_3)^3 = (x_1^3+x_2^3+x_3^3) + 3(x_1^2x_2+x_1^2x_3+x_1x_2^2+x_2^2x_3+x_1x_3^2+x_2x_3^2) + 6x_1x_2x_3$. After that I expanded $(x_1x_2+x_1x_3+x_2x_3)^2 = 2(x_1x_2x_3)+x_1^2x_2^2+x_1^2x_3^2+x_2^2x_3^2$. I don't know if my way of doing it is the right way, and if it is can you help from this point on?

Use Newton's identities https://en.wikipedia.org/wiki/Newton%27s_identities — Jean Marie, Jun 12 '19 at 16:09
It should be an expression that include m. I don't understand why is 3. — Raducu Mihai, Jun 12 '19 at 16:09

Archis Welankar · Answer 1 · 2019-06-12T16:26:46.450

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Since $x_1,x_2,x_3$ are the roots we find that $x_1^3=3x_1-mx_1^2-1$ similarly for $x_2,x_3$ adding the three equations we find that $x_1^3+x_2^3+x_3^3=3(x_1+x_2+x_3)-m(x_1^2+x_2^2+x_3^2)-3$ we see that $(x_1+x_2+x_3)^2=x_1^2+x_2^2+x_3^2+2(x_1x_2+x_2x_3+x_1x_3)$ thus the required value is $x_1^3+x_2^3+x_3^3=3(-m)-m(((-m)^2)-2(-3))-3=-3m-m^3-6m-3=-m^3-9m-3$

edited Jun 12 '19 at 16:26

answered Jun 12 '19 at 16:14

Archis Welankar

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$-m^3-9m-3$ I think. – Nosrati Jun 12 '19 at 16:20
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You should have $x_1+x_2+x_3=-m.$ – Thomas Andrews Jun 12 '19 at 16:21
You should also have $x_1x_2+x_2x_3+x_1x_3=-3.$ You have $3.$ – Thomas Andrews Jun 12 '19 at 16:25

score 1 · Accepted Answer · answered Jun 12 '19 at 16:18

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We have that $x^3_1+x^3_2+x^3_3$ is symmetric (fixed by any element of $S_3$).

Therefore, you may express $x^3_1+x^3_2+x^3_3$ using only the elementary symmetric polynomials.

You may use symmetric reduction on wolframalpha to deduce that:

$x^3_1+x^3_2+x^3_3=(x_1+x_2+x_3)^3-3(x_1x_2+x_2x_3+x_1x_3)(x_1+x_2+x_3)+3x_1x_2x_3$.

Now plug in to get $x^3_1+x^3_2+x^3_3$ in terms of the coefficients.

I believe you should get $(-m)^3-3(-3)(-m)+3(-1)=-m^3-9m-3$

answered Jun 12 '19 at 16:18

Locally unskillful

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I used symmetric polynomials, but the result shouldn't be $-m^3-6m^2+9m$? – Raducu Mihai Jun 12 '19 at 16:26
In your original post you have $x_1+x_2+x_3=m $ instead of $-m$, $x_1x_2+x_2x_3+x_1x_3=3$ instead of $-3$ and $x_1x_2x_3=1$ instead of $-1$. Is that why? – Locally unskillful Jun 12 '19 at 16:28
Yeah, I think so. Anyways symmetric polynomials are what I need. Thank you for the help. – Raducu Mihai Jun 12 '19 at 16:33
May I ask you another question about this exercise ?I don't know if it is a good idea to post a new question. – Raducu Mihai Jun 12 '19 at 16:38

Find $x_1^3+x_2^3+x_3^3$ for a degree 3 polynomial.

2 Answers2