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
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cube root of (x + y^2)
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sanju09
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scoobydooby
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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
(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
