cube root of (x + y^2)

[sanju09]

If x and y are positive integers, is the cube root of (x + y^2) an integer?


(1) x = y^2 multiplied to (y-1)

(2) x = 2

[scoobydooby]

1) x=y^3-y^2
(x + y^2)^1/3=>(y^3-y^2+y^2)^1/3=(y^3)^1/3=y an integer (given)
sufficient

2) x = 2

if y=2, (2+2^2)^1/3=(6)^1/3 not an integer as 6 is no perfect cube
if y=5, (2+5^2)^1/3=(27)^1/3=3 integer
not sufficient

hence A