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# M01-12

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Math Expert
Joined: 02 Sep 2009
Posts: 51296

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15 Sep 2014, 23:15
00:00

Difficulty:

25% (medium)

Question Stats:

73% (00:32) correct 27% (00:39) wrong based on 165 sessions

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What is the value of $$p^3 - q^3$$?

(1) $$p - q = 0$$

(2) $$p + q = 0$$

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Math Expert
Joined: 02 Sep 2009
Posts: 51296

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15 Sep 2014, 23:15
2
Official Solution:

(1) $$p-q=0$$. Rearrange: $$p=q$$, so $$p^3-q^3=p^3-p^3=0$$. Sufficient.

(2) $$p+q=0$$. Rearrange: $$q=-p$$, so $$p^3-q^3=p^3-(-p^3)=2p^3$$, so the value we are looking for depends on the value of $$p$$. Not sufficient.

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Joined: 08 Apr 2014
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Location: United States
Concentration: Strategy, Technology
GPA: 4

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05 Feb 2015, 05:27
Hi Bunuel,
I solved the question in the following way.
1)p-q=0;
(p-q)^3=0;
p^3-q^3-3p^2q-3pq^2=0;
p^3-q^3-3pq(p-q)=0;
//But we know that p-q=0
p^3-q^3=0//Sufficient

2)p+q=0;
From this we will get p^3+q^3 which is not sufficient.

Is this approach correct?

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Location: Hungary
Schools: Erasmus '19
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02 May 2017, 07:13
I, on the other hand, changed the question stem to "is p^3=q^3?"

I believe it's easier to tackle the question like this.
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"When the going gets tough, the tough gets going!"

|Welcoming tips/suggestions/advices (you name it) to help me achieve a 600|

Manager
Joined: 19 Aug 2016
Posts: 149
Location: India
GMAT 1: 640 Q47 V31
GPA: 3.82

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20 Jul 2017, 04:53
Bunuel wrote:
Official Solution:

(1) $$p-q=0$$. Rearrange: $$p=q$$, so $$p^3-q^3=p^3-p^3=0$$. Sufficient.

(2) $$p+q=0$$. Rearrange: $$q=-p$$, so $$p^3-q^3=p^3-(-p^3)=2p^3$$, so the value we are looking for depends on the value of $$p$$. Not sufficient.

Hi,

Is expanding the equation $$p^3 - q^3$$ as$$(p-q)$$ $$p^2 - q^2$$wrong? From here we can use the $$a^2 - b^2$$exapnsion and solve.

Kindly explain!
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Math Expert
Joined: 02 Sep 2009
Posts: 51296

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20 Jul 2017, 05:01
ashikaverma13 wrote:
Bunuel wrote:
Official Solution:

(1) $$p-q=0$$. Rearrange: $$p=q$$, so $$p^3-q^3=p^3-p^3=0$$. Sufficient.

(2) $$p+q=0$$. Rearrange: $$q=-p$$, so $$p^3-q^3=p^3-(-p^3)=2p^3$$, so the value we are looking for depends on the value of $$p$$. Not sufficient.

Hi,

Is expanding the equation $$p^3 - q^3$$ as$$(p-q)$$ $$p^2 - q^2$$wrong? From here we can use the $$a^2 - b^2$$exapnsion and solve.

Kindly explain!

$$p^3 - q^3=(p - q) (p^2 + p q + q^2)$$, not $$(p-q)(p^2 - q^2)$$.
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Intern
Joined: 02 Jun 2015
Posts: 22
Location: United States

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03 Apr 2018, 14:34
To make my life easier I used Picking Numbers for P^3 - Q^3 = ?

Statement 1:
P - Q = 0

P = Q

If P = 1 Q = 1 then 1^3 - 1^3 = 1 - 1 = 0

If P = 2 Q = 2 then 2^3 - 2^3 = 8 - 8 = 0

Sufficient.

Statement 2:
P + Q = 0

P = - Q

If P = -1 Q = 1 then -1^3 - 1^3 = - 1 - 1 = -2

If P = -2 Q = 2 then -2^3 - 2^3 = -8 - 8 = -16

Thanks! Ale
Intern
Joined: 04 Aug 2018
Posts: 6

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01 Sep 2018, 10:45
Trick :-
(p-q) is a factor of the given statement. Since p & q both have odd number of powers.
(p+q) will be factor only if p & q are raised to even numbers of powers. for e.g. $$p2−q2$$
Re: M01-12 &nbs [#permalink] 01 Sep 2018, 10:45
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# M01-12

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