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If p^3 = -r^3, is p < r ?

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If p^3 = -r^3, is p < r ?

by gmattesttaker2 » Sun Mar 02, 2014 12:52 am
Hello,

I had a question here in Statement 1 :

If p^3 = -r^3, is p < r ?

(1) r = 2
(2) p^2 = 4

[spoiler]OA: A[/spoiler]


I tried to solve as follows:

1) r = 2

Given, p^3 = -r^3
=> p^3 = -2^3
=> p^3 = -8
=> p = (-8)^(1/3)
=> p = (-1)^(1/3) x 8^(1/3)

Now when we have (-1)^(1/3) will this be -1 ? I was wondering whether -1 raised to any number between 0 and 2 will be -1 only. Thanks for your help.

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Source: — Data Sufficiency |
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by GMATGuruNY » Sun Mar 02, 2014 4:18 am
gmattesttaker2 wrote:Hello,

I had a question here in Statement 1 :

If p^3 = -r^3, is p < r ?

(1) r = 2
(2) p^2 = 4

[spoiler]OA: A[/spoiler]
p³ = -r³ implies that p and r are DIFFERENT SIGNS.

Statement 1: r=2
Here, r>0, implying that p<0.
Thus, p<r.
SUFFICIENT.

Statement 2: p²=4
If p=-2, then r is POSITIVE, in which case p<r.
If p=2, then r is NEGATIVE, in which case p>r.
INSUFFICIENT.

The correct answer is A.

In a fractional exponent, the numerator represents a POWER, while the denominator represents a ROOT.
(-1)^(1/3) = the CUBE ROOT of -1 raised to a POWER of 1.
Thus:
(-1)^(1/3) =∛(-1)¹ = ∛(-1) = -1.
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