Each of x, y and z is a rational number such that: x + y * ³√2 + z * ³√4 = 0.

Prove that x=y=z=0 is the only possible solution to the above equation.

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