Question

Heres a problem for fun, given a and b, solve this system of equations:
x + y + z = 0
x^2 + y^2 +z^2 = 6ab
x^3 + y^3 + z^3 = 3(a^3+b^3)
I'll post a solution sometime tonight or tomorrow or whatever if people want to see it.

Answer 1

just so it looks a little nicer:
\[x+y+z = 0\]
\[x^{2}+y^{2}+z^{2} = 6ab\]
\[x^{3}+y^{3}+z^{3} = 3(a^{3}+b^{3})\]
a and b are fixed.

Answer 2

See the attachment.

Answer 3

Yep that is correct :) Here's a pdf of my solution. The idea for my solution comes from this concept:
Lets say you have a system of equations:
\[a +b = 8\]
\[ab = 15\]
I want to know of 2 numbers that add up to 8, and multiply up to 15. There is symmetry in this problem (if I knew one solution, say a =x and b = y, then i can switch them and it would be another solution). Because of the symmetry, asking someone to solve this system of equations is the exact same thing as asking someone to solve this:
\[x^{2}-8x+15 = 0\]
Now why is that? It is because if I knew the 2 roots of a quadratic equation, say r and s, then:
\[(x-r)(x-s) = x^{2}-(r+s)x+(rs)\]
The coefficients of a quadratic polynomial are the sum and product of its roots. This idea can be extended to higher degree polynomials, and thats basically what I did. I created a cubic polynomial that is equivalent to the system, and solved the polynomial instead.