[quote="solarzj"]If x and y are positive integers, is the following cube root an integer?
(x + y^2)^(1/3)
1. x = y^2(y-1)
2. x = 2
So if I understand correctly.
In 1.) if I expand x = y^2 * (y-1) to x = y^3 - y ^2 and substitute in
cubeRoot(x + y^2) => i get cubeRoot(y^3 - y ^2 + y^2)
which leads to cubeRoot(y^3)
because both y^2 cancel out.
After this I can say, based on the information of the question (x and Y are positive integers) that y is a positive INTEGER
and because cubeRoot(y^3) => y ... the answer is yes, cubeRoot(x + y^2) = y = a positive INTEGER
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