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How to prove that ${x_1^3+x_2^3 + x_3^3 = 3x_1x_2x_3}$ holds in case ${x_1, x_2, x_3}$ are roots of the polynomial?

I've tried the following approach:

If $x_1$, $x_2$ and $x_3$ are roots then

$$(x-x_1)(x-x_2)(x-x_3) = x^3+px+q = 0$$

Now find the coefficient near the powers of $x$:

$$ x^3 - (x_1 + x_2 + x_3)x^2 + (x_1x_2 + x_1x_3 + x_2x_3)x - x_1x_2x_3 = x^3+px+q $$

That means that I can write a system of equations:

$$ \begin{cases} -(x_1 + x_2 + x_3) = 0 \\ x_1x_2 + x_1x_3 + x_2x_3 = p \\ - x_1x_2x_3 = q \end{cases} $$

At this point I got stuck. I've tried to raise $x_1 + x_2 + x_3$ to 3 power and expand the terms, but that didn't give me any insights. It feels like I have to play with the system of equations in some way but not sure what exact.

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    $\begingroup$ If you ever need to go further, I recommend that you study Newton's identities. They settle this question also. But, as shown by the answers below, you don't need their full power to reach the sum of cubes. $\endgroup$ – Jyrki Lahtonen Jun 11 '18 at 18:32
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Rewrite $x_1,x_2,x_3$ with $a,b,c$. From first Vieta formula we have $$a+b+c=0$$ so $a+b=-c$ and so on...

Now $$a^3+b^3+c^3= (a+b)(a^2-ab+b^2)+c^3 = c(\underbrace{-a^2+ab-b^2+c^2}_I)$$

Since

$$I = -a^2+ab-b^2+c^2 = a(b-a)+(c-b)(c+b) = $$ $$a(b-a)-a(c-b) = a(2b-a-c)=a(2b+b)=3ab$$

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$x_1^3 + x_2^3 + x_3^3 - 3x_1x_2x_3 = (x_1 + x_2 + x_3)(x_1^2 + x_2^2 + x_3^2-x_1x_2 - x_2x_3-x_1x_3)$

Now, here $-(x_1 + x_2 + x_3) =$ Coefficient of $x^2$/ Coefficient of $x^3 =0 $

So, $x_1^3 + x_2^3 + x_3^3 = 3x_1x_2x_3$

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Hint: $\,x_1^3+px_1+q=0 \iff x_1^3=-px_1-q\,$, then adding the three relations together:

$$\require{cancel} x_1^3+x_2^3+x_3^3=\cancel{-p(x_1+x_2+x_3)}-3q=\cdots $$

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    $\begingroup$ Great solution +1 $\endgroup$ – Aqua Jun 11 '18 at 17:30
  • $\begingroup$ @ChristianF Thanks. One advantage of this approach is that it easily extends to equations $\,x^{\color{red}{n}}+px+q=0\,$, where $\,\sum_i x_i^n = (-1)^{n+1}n \prod_i x_i\,$. $\endgroup$ – dxiv Jun 11 '18 at 21:13
  • $\begingroup$ @dxiv I understand this method is one of the basic ones used all over the places, but is there some source where I can read about such methods. Jyrki Lahtonen has mentioned Newton's Identities above, is it something that comes from there? $\endgroup$ – roman Jun 12 '18 at 5:08
  • $\begingroup$ @RomanKapitonov My shortcut takes advantage of the particular form of the given polynomial. In general, the elementary symmetric polynomials, Vieta's relations and Newton's identities should get you through. $\endgroup$ – dxiv Jun 12 '18 at 5:40
  • $\begingroup$ @dxiv Thank you for your guidance $\endgroup$ – roman Jun 12 '18 at 9:40
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If $ x_1,x_2,x_3 $ are roots of $ x^3+p x+q=0 $ then $ x_1+x_2+x_3 = 0 $

If $ x_1+x_2+x_3 = 0 $ then $ x_3 = -(x_1+x_2) $ and

$ x_1^3+x_2^3+x_3^3 = x_1^3+x_2^3+(-1)^3(x_1+x_2)^3 = -3(x_1^2x_2+x_1x_2^2) = -3x_1x_2(x_1+x_2) = -3x_1x_2(-x_3) = 3x_1x_2x_3 $

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Every symmetric polynomial can be expressed in terms of the elementary symmetric polynomials, in this case $s_1=x_1+x_2+x_3$, $s_2=x_1x_2+x_2x_3+x_3x_1$ and $s_3=x_1x_2x_3$. Since $x_1^3+x_2^3+x^3$ is homogeneous, we can find $a$, $b$ and $c$ such that $$ x_1^3+x_2^3+x_3^3=as_1^3+bs_1s_2+cs_3 $$

  • For $x_1=1$, $x_2=0$, $x_3=0$: $1=a$
  • For $x_1=1$, $x_2=1$, $x_3=0$: $2=8a+2b$
  • For $x_1=1$, $x_2=1$, $x_3=1$: $3=27a+9b+c$

Therefore $a=1$, $b=-3$, $c=3$ and finally $$ x_1^3+x_2^3+x_3^3=(x_1+x_2+x_3)^3-3(x_1+x_2+x_3)(x_1x_2+x_2x_3+x_3x_1)+3x_1x_2x_3 $$ This is a general result.

In your case, by Viète’s formulas $$ x_1+x_2+x_3=0,\qquad $$ so in the end $$ x_1^3+x_2^3+x_3^3=3x_1x_2x_3 $$

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