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

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?

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 $$(-m)^3-3(-3)(-m)+3(-1)=-m^3-9m-3$$

• 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

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 $$x_1^3+x_2^3+x_3^3=3(-m)-m(((-m)^2)-2(-3))-3=-3m-m^3-6m-3=-m^3-9m-3$$

• $-m^3-9m-3$ I think. – Nosrati Jun 12 '19 at 16:20
• 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