Using r to denote the cube root of 2, it follows that cube root of 4 is r^{2} and our equation can be written as:
x + ry + r^{2}z = 0 (1)
Rearranging and squaring gives x^{2} = r^{2}y^{2} + 2r^{3}yz + r^{4}z^{2} and, using the fact that
r^{3} = 2 and r^{4} = 2r, this can be written as
x^{2} = r^{2}y^{2} + 4yz +2rz^{2} (2)
Eliminating the term involving r^{2} between (1) and (2) now gives
xy^{2} + x^{2}z = ry^{3} + 4yz^{2} +2rz^{3}
xy^{2} + x^{2}z  4yz^{2} = r(2z^{3}  y^{3})
If each side is nonzero, then, since r is irrational, x, y and z cannot be rational.
If each side is zero, then y = (cube root 2) z, which also denies y and z being rational unless y = z = 0.
Using (1), it follows that x = y = z = 0.
(Still can't get the extra font facility to give me a root sign..)
Edited on October 15, 2009, 7:58 pm
Edited on October 15, 2009, 8:02 pm

Posted by Harry
on 20091015 19:55:35 