I have we have to solve this intimidating system in R:
x+y+z=3
x^3+y^3+z^3=3
x^5+y^5+z^5=3
I tried unsuccessfully to define a polynomial with solutions x y z but i dont have enough relationships to tie something.
Can u guide me to a ideea ,especialy about x^5+y^5+z^5
[MATH]\text {DEFINE: } f(x, y, z) = 3 - (x + y + z),\ g(x, y, z) = 3 - (x^3 + y^3 + z^3), \text { and } h(x, y, z) = 3 - (x^5 + y^5 + z^5).[/MATH]
The system in the problem above has three equations and three unknowns. In principle, the system
DOES have sufficient relationships to generate a positive number of solutions if the equations are
independent and consistent. It is easy enough to show that they are independent, but showing that they are consistent is most easily done, perhaps only feasible, by showing that there is at least one solution.
One such solution can be found by symmetry, namely (1, 1, 1), thereby proving consistency. But that does not show that the solution is unique. Moreover, someone might not "see" the symmetry solution. Here is a method to find all the solutions.
Notice first that any solution to any of the equations is a zero of one of the functions that I have defined. Of course, there are an infinite number of zeroes of each function, but a solution to the system must be a point that is a zero of all three functions. The distance between the functions at a solution must be zero. Thus, the set of points where d = 0 will contain the solution
if one exists.
The distance between f and g at an arbitrary point (p, q, r) is
[MATH]d = \sqrt{(p^3 - p)^2 + (q^3 - q)^2 + (r^3 - r)^2}.[/MATH]
Notice that the addends in the square root are all non-negative. Therefore d = 0 entails that each addend be zero.
[MATH](p^3 - p)^2 = 0 \implies \implies p = p^3 \implies p = -\ 1, 0, \text { or } 1.[/MATH]
[MATH]d = 0 \implies (0,\ 0,\ 0),\ (0,\ 0,\ \pm 1),\ (0,\ \pm 1,\ 0),\ (0,\ \pm 1,\ \pm1),[/MATH]
[MATH](\pm 1,\ 0,\ 0),\ (\pm 1,\ 0,\ \pm1),\ (\pm 1,\ \pm 1,\ 0),\ \text { or } (\pm 1,\ \pm 1,\ \pm1).[/MATH]
Only one of those points is a zero either f or g. (1, 1, 1) is a zero of both f and g. Furthermore, (1, 1, 1) is a zero of h. There is a unique solution, namely (1, 1, 1).
