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The cubic equation

$$y^3 - y^2 - 4y + 3 = 0$$

has real solutions $y_1$, $y_2$, and $y_3$. How do I create another cubic equation with integer coefficients that has solutions: $y_1 + y_2$, $y_1 + y_3$, and $y_2 + y_3$?

At first glance, it looked as if applying Viète's formulae and symmetric polynomials would be useful, but as I worked through it, I got a bit lost when it came to creating another cubic equation.

I have already solved for the solutions without using Viète's formulae and they are all irrational... I wrote the following equalities down to help with the problem:

\begin{align} y_1 + y-2 + y_3 &= 1\\ y_1y_2 + y_1y_3 + y_2y_3 &= -4\\ y_1y_2y_3 &= -3 \end{align}

I deduced these from Viète's formulae but I don't really know how to apply them to solve the problem. I did a bit more research and found that elementary symmetric polynomials are helpful with these kinds of questions too. Overall, I would really like to know how one could apply Viète's formulae, and theorems regarding elementary/symmetric polynomials to solve this problem as these concepts are quite new to me.

I've seen a problem like this on this site, but it includes finding polynomials where the roots are isolated to only one of the roots of the original equation (i.e., they are $\frac1{a_1}$,$\frac1{a_2}$,$\frac1{a_3}$ as opposed to $\frac1{a_1} + \frac1{a_2}$ etc.) It would be extremely appreciated if someone could assist me with this problem.


Update: I think I've solved for $p$: $$y_1 + y_2 + y_3 + y_1 + y_2 + y_3 = 2$$ Therefore, $p = -2$? However I'm still stumped on how to solve for $r$ and $q$...

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  • $\begingroup$ How do I use MathJax? Sorry, I'm not really familiar with this $\endgroup$ – Alexander B Jun 9 '19 at 2:39
  • $\begingroup$ Here is the page for MathJax and you enclose formulas in dollar signs (\$x^2\$). $\endgroup$ – Toby Mak Jun 9 '19 at 2:40
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As you've mentioned, you know that $$y_1 + y_2 + y_3 = 1$$ $$y_1y_2 + y_1y_3 + y_2y_3 = -4$$ $$y_1y_2y_3 = -3$$

To find a polynomial $x^3 - ax^2 + bx - c$ whose roots are $y_1 + y_2, y_1 + y_3$ and $y_2 + y_3$, you can use Vieta's formulas in terms of elementary symmetric polynomials of these roots. That is, you want to calculate:

$$a = (y_1 + y_2) + (y_1 + y_3) + (y_2 + y_3)$$ $$b = (y_1 + y_2)(y_1 + y_3) + (y_1 + y_2)(y_2 + y_3) + (y_1 + y_3)(y_2 + y_3)$$ $$c = (y_1 + y_2)(y_1 + y_3)(y_2 + y_3)$$

The first one is easy to find, it is equal to $a = 2(y_1 + y_2 + y_3) = 2$.

The computations for $b$ and $c$ are more tedious, but they are symmetric polynomials in $y_1, y_2$ and $y_3$ so they can be expressed in terms of elementary symmetric polynomials (and hence in terms of known values) using something like Gauss' algorithm.

In our case we have $b = (y_1+y_2+y_3)^2 + (y_1y_2 + y_1y_3 + y_2y_3) = -3$ and we have $c = (y_1 + y_2 + y_3)(y_1y_2 + y_1y_3 + y_2y_3) - y_1y_2y_3 = -1$.

So the polynomial $x^3 - 2x^2 - 3x + 1$ should have the right roots, assuming my calculations are correct.

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  • $\begingroup$ How did you get to (y1+y2+y3)^2? I couldn't factor (y1)^2 + (y2)^2 + (y3)^2 $\endgroup$ – Alexander B Jun 9 '19 at 3:18
  • $\begingroup$ Where does $y_1^2 + y_2^2 + y_3^2$ come from? Also, as I mentioned, the general method to write symmetric polynomials in terms of elementary symmetric polynomials relies on Gauss' algorithm. The idea is that you can order the monomials in a way that depends on the degree of each variable, and then you can select appropriate products of elementary symmetric polynomials to subtract. This allows you to reduce the "multi-degree" of your polynomial, and the algorithm terminates by induction. The details can be found in proofs of the Fundamental Theorem of Symmetric Polynomials. $\endgroup$ – Tob Ernack Jun 9 '19 at 3:20
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By Vieta’s formula, we know that $y_1+y_2+y_3 = 1$. So we’re looking for the cubic whose roots are $1-y_1$, $1-y_2$, $1-y_3$. This is accomplished by doing the simple substitution (noting that the inverse function of $1-x$ is itself): $$f(1-y) = (1-y)^3 - (1-y)^2 - 4(1-y) + 3 = -y^3 + 2y^2 + 3y - 1.$$

Negating this to make it monic yields $y^3 - 2y^2 - 3y + 1$, as in Tob Ernack’s answer.

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  • 2
    $\begingroup$ Wow this is quite nicer! $\endgroup$ – Tob Ernack Jun 9 '19 at 16:49
  • $\begingroup$ @TobErnack Thank you! I appreciate that your answer provides a general framework for approaching all similar such problems. $\endgroup$ – Erick Wong Jun 9 '19 at 17:01

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