Using r to denote the cube root of 2, it follows that cube root of 4 is r2 and our equation can be written as:
x + ry + r2z = 0 (1)
Rearranging and squaring gives x2 = r2y2 + 2r3yz + r4z2 and, using the fact that
r3 = 2 and r4 = 2r, this can be written as
x2 = r2y2 + 4yz +2rz2 (2)
Eliminating the term involving r2 between (1) and (2) now gives
xy2 + x2z = -ry3 + 4yz2 +2rz3
xy2 + x2z - 4yz2 = r(2z3 - y3)
If each side is non-zero, 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 2009-10-15 19:55:35