Prove that there does not exist any triplet (F, G, H) of rational numbers, that satisfies this equation.

F + G*√H = ³√2

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)=2-f^3-f*g^2*h

sqrt(h)=(2-f^3-f*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 2009-07-23 12:34:12