# How to create a cubic equation that include sums of the roots of another cubic equation

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 $$q$$...

• How do I use MathJax? Sorry, I'm not really familiar with this – Alexander B Jun 9 '19 at 2:39
• Here is the page for MathJax and you enclose formulas in dollar signs (\$x^2\$). – Toby Mak Jun 9 '19 at 2:40

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.

• How did you get to (y1+y2+y3)^2? I couldn't factor (y1)^2 + (y2)^2 + (y3)^2 – Alexander B Jun 9 '19 at 3:18
• 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. – Tob Ernack Jun 9 '19 at 3:20

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.

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