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cube root of (x + y^2)

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by sanju09 » Sat Apr 04, 2009 6:09 am
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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Source: — Data Sufficiency |
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by scoobydooby » Sun Apr 05, 2009 4:11 am
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
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