Show that if x, y, z are integers such that x^3+5y^3 = 25z^3, then x = y = 0.

3125*k^9 + y^3 is an integer my closure property.

 but 5^(1/3) is not an integer, which forces z to be irrational.

Note that there is no way an integer value can rationalize 5^(1/3)

Step-by-step explanation:

x^3 = 25z^3 - 5y^3

x^3 = 5 ( 5z^3 - y^3)

x  =  (5 ( 5z^3 - y^3) )^(1/3) must be an integer

  = 5^(1/3) * (5z^3 - y^3)^(1/3)

Then  (5z^3 - y^3)^(1/3) = 25*k^3 for some integer k

 5z^3 - y^3 = 15625*k^9

 5z^3 = 15625*k^9 + y^3

  z^3 = 3125*k^9 + (1/5)*y^3

z =  ( 3125*k^9 + (1/5)*y^3 )^ ( 1/3)