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)