Finding the Roots of a CubicDate: 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 |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/