x=c-y-z
b=(c-y-z)y + (c-y-z)z + yz = c(y+z)-y^2-zy-z^2
Solve this quadratic equation for y (or z), use both in a=xyz and hope that it has a nice solution?
#3
EM_Guy
217
49
It is not a quadratic equation. And it is not a "nice" solution.
I have determined that z^3-cz^2+bz-a = 0. So, if we can find the roots of the cubic function, then we have z as a function of a, b, and c. Then, it should be straightforward to find x and y in terms of a, b, and c.
But I forget how to find the roots of a cubic function.
It is well known that a vector space always admits an algebraic (Hamel) basis. This is a theorem that follows from Zorn's lemma based on the Axiom of Choice (AC).
Now consider any specific instance of vector space. Since the AC axiom may or may not be included in the underlying set theory, might there be examples of vector spaces in which an Hamel basis actually doesn't exist ?