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# If x and y are positive integers, is the following cube root

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If x and y are positive integers, is the following cube root [#permalink]

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09 Jun 2012, 07:22
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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
[Reveal] Spoiler: OA

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Intern
Joined: 14 Mar 2012
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Re: If x and y are positive integers, is the following cube root [#permalink]

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09 Jun 2012, 07:23
[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)
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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Re: If x and y are positive integers, is the following cube root [#permalink]

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09 Jun 2012, 07:45
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If x and y are positive integers, is the following cube root an integer?

$$\sqrt[3]{x + y^2}$$

(1) x = y^2(y-1) --> $$x=y^3-y^2$$ --> $$\sqrt[3]{x+y^2}=\sqrt[3]{y^3-y^2+y^2}=\sqrt[3]{y^3}=y=integer$$. Sufficient.

(2) x = 2 --> if $$y=1$$ then the answer is NO but if $$y=5$$ then the answer is YES. Not sufficient.

Hope it's clear.
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Re: If x and y are positive integers, is the following cube root [#permalink]

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27 Aug 2015, 08:22
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If x and y are positive integers, is the following cube root   [#permalink] 27 Aug 2015, 08:22
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