 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  x + y * 3√2 + z * 3√4 = 0 (Posted on 2009-10-15) 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) Comments: ( Back to comment list | You must be logged in to post comments.) Solution - trying formatting | Comment 10 of 14 | I'm using the same strategy as Harry in my solution, with a generalization, let c be any integer which is not a perfect cube (non-zero is covered by not perfect cube).

Given Equation
(1) x + ����(c)*y + ����(c��)*z = 0

Rearrange and square (1)
(2) x�� = ����(c��)*y�� + 2c*yz +c*����c*z��

Equation 2 multiplied by z and rearranged
(3) x��*z = (����(c��)*z)*y�� + 2c*yz�� +c*����c*z��

Substitute (1) into (3) to eliminate (����c^2*z)
(4) x��*z = (-x - ����c*y)*y�� + 2c*yz�� +c*����c*z��

Rearrange (4) according to terms with cbrt(c)
(5) x��*z + y��*z - 2*c*y*z�� = ����c*(c*z�� - y��)

The left side (x��*z + y��*z - 2*c*y*z��) is rational.  The right factor (c*z�� - y��) is rational, but ����c is irrational. The only way the equation is true is if the rational factors are zero.

Equations without formatting:
(1) x + cbrt[c]*y + cbrt[c^2]*z = 0
(2) x^2 = cbrt[c^2]*y^2 + 2c*yz +c*cbrt[c]*z^2
(3) x^2*z = (cbrt[c^2]*z)*y^2 + 2c*yz^2 +c*cbrt[c]*z^3
(4) x^2*z = (-x - cbrt[c]*y)*y^2 + 2c*yz^2 +c*cbrt[c]*z^3
(5) x^2*z + y^2*z - 2*c*y*z^2 = cbrt[c]*(c*z^3 - y^3)

 Posted by Brian Smith on 2009-10-16 12:04:25 Please log in:
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