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x + y * 3√2 + z * 3√4 = 0 (Posted on 2009-10-15) Difficulty: 3 of 5
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.

No Solution Yet Submitted by K Sengupta    
Rating: 3.0000 (1 votes)

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Solution Solution | Comment 3 of 14 |
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

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