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**1. Background/theory**

We know that if the equation x

^{3}+px

^{2}+qx+r=0 has solutions x

_{1}, x

_{2}, x

_{3}then

x

_{1}+ x

_{2}+ x

_{3}= -p

x

_{1}x

_{2}+ x

_{2}x

_{3}+ x

_{3}x

_{1}= q

x

_{1}x

_{2}x

_{3}= -r

**2. Problem statement**

Find (x

_{1}- x

_{2})

^{2}(x

_{2}- x

_{3})

^{2}(x

_{3}- x

_{1})

^{2}as an expression containing p,q,r.

That is, I'm supposed to find the discriminant of the above cubic equation in terms of its coefficients.

## The Attempt at a Solution

Now, I have approached the problem in two ways. First, I expanded the whole expression out and tried to manipulate the expression to get it in the desired terms. It is a very lenthy and laborious process though and I had trouble getting the right expression. So there must be some more elegant solution to this problem than just multiplying the expression out and manipulate it.

My best attempt at a "more elegant" solution was to reason as follows:

When we expand the expression (x

_{1}- x

_{2})

^{2}(x

_{2}- x

_{3})

^{2}(x

_{3}- x

_{1})

^{2}we get a polynomial with terms of degree 6 in x

_{1}, x

_{2}, x

_{3}. Now, p has degree 1, q has degree 2 and r has degree 6 so the possible terms in the expression is:

p

^{6}, q

^{3}, r

^{2}, pqr, p

^{3}r, p

^{4}q, p

^{2}q

^{2}.

So we can write:

(x

_{1}- x

_{2})

^{2}(x

_{2}- x

_{3})

^{2}(x

_{3}- x

_{1})

^{2}= Ap

^{6}+ Bq

^{3}+ Cr

^{2}+ Dpqr + Ep

^{3}r + Fp

^{4}q + Gp

^{2}q

^{2}where A,...,G are constants.

Then I tried to determine A,...,G by finding cubic polynomials with given roots and thus given values of the discriminant and p, q, r. I easily determined the terms -27r

^{2}and -4q

^{3}by just letting p=0 but when I tried to determine the other constants I got wierd numbers like A=770,4 and G=87,7 etc, whereas I expected those constants to be zero (since the discriminant does not contain p

^{2}q

^{2}, for instance). I can't explain why...

I am grateful for any guidance here! Also, I apologize for possible language errors, English is not my native.