Algebra PuzzleDate: 04/09/2001 at 18:26:47 From: Johann Matias H. L. Subject: Hard equation Okay, here you go... if this is true: x + y + z = 1 x*x + y*y + z*z = 2 x*x*x + y*y*y + z*z*z = 3 find (without solving the equation): x*x*x*x + y*y*y*y + z*z*z*z = ? I've tried everything. Please help me! Date: 04/16/2001 at 12:06:35 From: Doctor Peterson Subject: Re: Hard equation Hello, Johann. This question stuck in my mind until I had the time to solve it, and I want to give you a hint you can try. I will write "x^2" for x squared, and so on. I will also generalize the problem a little. Suppose we know that: x + y + z = A x^2 + y^2 + z^2 = B x^3 + y^3 + z^3 = C Now we want to find D: x^4 + y^4 + z^4 = D I tried combining the original equations in various ways to get something that includes D and other symmetrical expressions; here is one that I found useful: AC = (x + y + z)(x^3 + y^3 + z^3) = (x^4 + y^4 + z^4) + (xy^3 + xz^3 + yx^3 + yz^3 + zx^3 + zy^3) = (x^4 + y^4 + z^4) + (xy*x^2 + xy*y^2 + xy*z^2) + (yz*x^2 + yz*y^2 + yz*z^2) + (zx*x^2 + zx*y^2 + zx*z^2) - (xy*z^2 + yz*x^2 + zx*y^2) = (x^4 + y^4 + z^4) + (xy + yz + zx)*(x^2 + y^2 + z^2) - xyz*(x + y + z) = D + PB - QA where P = xy + yz + zx Q = xyz The fact that P and Q are symmetrical with respect to the three variables suggests that they might appear in other expressions involving A, B, and C; if we can find P and Q in terms of A, B, and C, we will be able to find D. To do this, I used: A^2 = (x + y + z)(x + y + z) = (x^2 + y^2 + z^2) + 2(xy + yz + zx) = B + 2P and AB = (x + y + z)(x^2 + y^2 + z^2) = (x^3 + y^3 + z^3) + (xy^2 + xz^2 + yx^2 + yz^2 + zx^2 + zy^2) = (x^3 + y^3 + z^3) + (xy*x + xy*y + xy*z - xyz) + (yz*x + yz*y + yz*z - xyz) + (zx*x + zx*y + zx*z - xyz) = (x^3 + y^3 + z^3) + xy*(x + y + z) + yz*(x + y + z) + zx*(x + y + z) - 3xyz = (x^3 + y^3 + z^3) + (xy + yz + zx)*(x + y + z) - 3xyz = C + PA - 3Q You'll be able to solve this for D. What I'm not sure of is whether there are any actual solutions in x, y, and z at all; but if there are, this will tell you what x^4 + y^4 + z^4 has to be. (I think actually solving for x, y, and z requires solving a sixth-degree polynomial, and they may be complex.) I'd feel a lot surer of my solution if I could check it! - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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