Solving x^3 + y^3 + z^3
Date: 02/22/2003 at 08:06:53 From: Martin Subject: Solving x^3 + y^3 + z^3 Given: x + y + z = 0; x * y * z = 2; x,y,z are not equal to each other; x,y,z all real numbers. Solve for x^3 + y^3 + z^3 When doing the algebra, what to do with the terms with x^2, y^2, or z^2 ? I start off with (x + y + z) = 0; therefore (x + y + z)^3 = 0. Multiplying out yields x^3 + y^3 + z^3 + 3x^2y + 3x^2z + 3y^2x + 3y^2z + 3z^2x + 3z^2y + 6x*y*z. I keep the x^3, y^3, and z^3 on the same side of the equals sign. I can then solve for their sum. On the other side, with ease I substitute 2 into the 6x*y*z = 6*2. The problem is what to do with the terms such as x^2y or y^2x, etc. I have tried every way of factoring hoping to be left with something I can work with. I also tried substituting in from (x + y + z) = 0 and x * y * z = 2. The factoring and substitution did not yield anything I can use. Is there a trick to dealing with the terms such as y^2z or did I take the wrong approach from the beginning?
Date: 02/22/2003 at 15:29:18 From: Doctor Greenie Subject: Re: Solving x^3 + y^3 + z^3 Hi, Martin - (1) Your approach to the whole problem is exactly on target. (2) Yes, there IS a trick - a really cool trick! - for dealing with the terms such as x^2y. We have 0 = (x+y+z)^3 = (x^3+y^3+z^3)+3(x^2y+x^2z+xy^2+y^2z+xz^2+yz^2)+6xyz and so x^3+y^3+z^3 = -3(x^2y+x^2z+xy^2+y^2z+xz^2+yz^2) - 6xyz = -3(x^2y+x^2z+xy^2+y^2z+xz^2+yz^2) - 12 Now our task is to evaluate the expression x^2y+x^2z+xy^2+y^2z+xz^2+yz^2 To do this, let's start by grouping terms and factoring so that there are no "squared" terms left: x^2y+x^2z+xy^2+y^2z+xz^2+yz^2 = x(xy+xz) + y(xy+yz) + z(xz+yz) And here comes the trick.... In this factored form, if we could get the three expressions in parentheses to be identical, then we would have x(...) + y(...) + z(...) = (x+y+z)(...) = 0(...) = 0 So let's add some terms (and then subtract those same terms) to do exactly that: x^2y+x^2z+xy^2+y^2z+xz^2+yz^2 = x(xy+xz) + y(xy+yz) + z(xz+yz) = x(xy+xz+yz-yz) + y(xy+yz+xz-xz) + z(xz+yz+xy-xy) = x(xy+xz+yz) + y(xy+xz+yz) + z(xy+xz+yz) - x(yz) - y(xz) - z(xy) = (x+y+z)(xy+xz+yz) - 3xyz = 0(xy+xz+yz) - 3(2) = 0 - 6 = -6 And then we are done: x^3+y^3+z^3 = -3(x^2y+x^2z+xy^2+y^2z+xz^2+yz^2) - 6xyz = -3(x^2y+x^2z+xy^2+y^2z+xz^2+yz^2) - 12 = -3(-6) - 12 = 18 - 12 = 6 I hope all this helps. Please write back if you have any further questions about any of this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/
Date: 02/22/2003 at 16:35:53 From: Martin Subject: Thank you (solving x^3 + y^3+z^3) Dr. Greenie, In the trick in the factoring that you demonstrated to me and I missed, it is evident what a beautiful subject math is! Thank you very much for your trouble. Both you and the whole Dr. Math group should NEVER be taken for granted. Best wishes, Martin
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