Finding the Roots of a Cubic

Date: 1/27/96 at 9:33:0
From: Anonymous
Subject: Finding roots of a cubic in x and y

Hi there. A friend at work was looking back at his old Math
exam papers, and came across a question which he couldn't see
how to solve.  He asked me to have a go, since I have a Math
degree. However since it was ten years ago I can't remember
much! 

The question is this:

Find all the roots of x + y = x^2 + y^2 = x^3 + y^3

My question is - what's the best approach for solving it?
It looks easy, and I'm sure it is. I've started off putting

k = x + y
k = x^2 + y^2
k = x^3 + y^3

etc., and expressing it in terms of k, but I still don't get
anywhere. Help!!!

(As far as I know there aren't intended to be any 'tricks' to it)

Date: 5/31/96 at 15:8:41
From: Doctor Alain
Subject: Re: Finding roots of a cubic in x and y

Let's write x + y = x^2 + y^2 = x^3 + y^3 in 2 equations

(i)   x + y = x^2 + y^2
(ii)  x + y = x^3 + y^3

(x^3 + y^3) can be factored so we should do it. Clearly if x = -y
then x^3 = -y^3 so (x+y) is a factor of (x^3 + y^3). After division
we get x^3 + y^3 = (x+y)(x^2-xy+y^2). This with (ii) gives

x + y = (x + y) (x^2 - xy + y^2),

so (A) x^2 - xy + y^2 = 1  OR  (B) x + y = 0.

(A) with (i) gives

x -xy + y = 1
x + (1-x)y = 1
(1-x)y = 1 - x

So (A.1) y = 1  OR (A.2) x = 1.

(A) with (A.1) gives x^2 - x = 0    x = 0  or 1.

First root (0,1)
Second root (1,1)

(A) with (A.2) gives y^2 - y = 0    y = 0 or 1.

Third root (1,0)
(1,1) is second root.

(B) with (i) gives x^2 + y^2 = 0 so x = y = 0.

Fourth root (0,0).

-Doctor Alain,  The Math Forum