I'm assuming that I can assume that the cube root of 2 is irrational and thus not have to prove it.
let
f+g*sqrt(h)=2^(1/3) cube both sides
f^3+f^2*g*sqrt(h)+f*g^2*h+g^3*h*sqrt(h)=2
(f^2*g+g^3*h)*sqrt(h)=2f^3f*g^2*h
sqrt(h)=(2f^3f*g^2*h)/(f^2*g+g^3*h)
now since all parts of the right hand side are rational then the right hand side is rational thus sqrt(h) must be rational.
If sqrt(h) is rational then so is f+g*sqrt(h) but that would mean that 2^(1/3) is rational which we know to not be true, thus there can not be any rational solutions to the equation
f+g*sqrt(h)=2^(1/3)

Posted by Daniel
on 20090723 12:34:12 