Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 02 Jul 2015, 04:48

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# What is the value of x^3 - y^3 ?

Author Message
TAGS:
Intern
Joined: 07 Feb 2009
Posts: 14
Followers: 0

Kudos [?]: 14 [1] , given: 13

What is the value of x^3 - y^3 ? [#permalink]  10 Oct 2010, 02:46
1
KUDOS
00:00

Difficulty:

45% (medium)

Question Stats:

56% (01:57) correct 44% (00:43) wrong based on 48 sessions
What is the value of x^3 - y^3 ?

(1) x^6 - y^6 = 0
(2) y = 0
[Reveal] Spoiler: OA
Retired Moderator
Joined: 02 Sep 2010
Posts: 805
Location: London
Followers: 83

Kudos [?]: 602 [1] , given: 25

Re: Powers doubt [#permalink]  10 Oct 2010, 02:54
1
KUDOS
vivaslluis wrote:
Hi,

I have a doubt with the following question:
- What is the value of x^3 - y^3 ?
(1) x^6 - y^6 = 0
(2) y = 0

I discard A, B and D because there's not enough information. I know the answer is C, but I don't get a clue of why.

Could someone help me with that?

Thank you!

$$x^6-y^6=0$$ and $$y=0$$
So $$x^6-0^6=0$$
So $$x=0$$
So $$x^3-y^3=0$$
_________________
Ms. Big Fat Panda
Status: Three Down.
Joined: 09 Jun 2010
Posts: 1923
Concentration: General Management, Nonprofit
Followers: 388

Kudos [?]: 1583 [2] , given: 210

Re: Powers doubt [#permalink]  10 Oct 2010, 10:12
2
KUDOS
Expert's post
vivaslluis wrote:
Hi,

I have a doubt with the following question: - What is the value of x^3 - y^3 ?
(1) x^6 - y^6 = 0
(2) y = 0

I discard A, B and D because there's not enough information. I know the answer is C, but I don't get a clue of why.

Could someone help me with that?

Thank you!

So you need $$x^3 - y^3$$.

Statement 1:

$$x^6 - y^6 = (x^3 - y^3)(x^3+y^3) = 0$$

Insufficient, since either one of the two terms could be zero.

Statement 2:

$$y^3 = 0$$, which means $$x^3 - y^3 = x^3.$$ Still insufficient.

Combining both, and substituting y = 0 in the second statement, we get : $$x^6 = 0$$, which means $$x = 0$$, and hence $$x^3 - y^3 = 0$$

Intern
Joined: 07 Feb 2009
Posts: 14
Followers: 0

Kudos [?]: 14 [1] , given: 13

Re: Powers doubt [#permalink]  10 Oct 2010, 13:47
1
KUDOS
Crystal clear. Thank you!
Manager
Joined: 08 Oct 2010
Posts: 213
Location: Uzbekistan
Schools: Johnson, Fuqua, Simon, Mendoza
WE 3: 10
Followers: 9

Kudos [?]: 348 [0], given: 974

Re: Powers doubt [#permalink]  23 Dec 2010, 08:06
I disagree with explanations given above. Here is my approach:

Stmt (1) indicates that x=y otherwise x^6-y^6=0 does not hold true. Hence, x^3-y^3 is always equal to zero. SUFF.

Stmt (2) is clearly INSUFF because there is no inf. about x.

If my approach is wrong, I would appreciate any other explanations.
Math Expert
Joined: 02 Sep 2009
Posts: 28242
Followers: 4461

Kudos [?]: 44999 [1] , given: 6638

Re: Powers doubt [#permalink]  23 Dec 2010, 08:30
1
KUDOS
Expert's post
feruz77 wrote:
I disagree with explanations given above. Here is my approach:

Stmt (1) indicates that x=y otherwise x^6-y^6=0 does not hold true. Hence, x^3-y^3 is always equal to zero. SUFF.

Stmt (2) is clearly INSUFF because there is no inf. about x.

If my approach is wrong, I would appreciate any other explanations.

Yes, your approach is wrong: $$x^6-y^6=0$$ implies that either $$x=y$$ or $$x=-y$$, for example $$1^6-1^6=0$$ and also $$1^6-(-1)^6=0$$. To see this algebraically you could rewrite $$x^6-y^6=0$$ as $$(x^3-y^3)(x^3+y^3)=0$$ --> either $$x^3=y^3$$, or $$x^3=-y^3$$ --> so either $$x=y$$ or $$x=-y$$. OR $$x^6-y^6=0$$ --> $$x^6=y^6$$ --> $$x^2=y^2$$ --> $$x=y$$ or $$x=-y$$.

Now, if $$x=y$$ then $$x^3-y^3=0$$ for any values of $$x$$ and $$y$$ BUT of $$x=-y$$ then $$x^3-y^3=2x^3$$ and we need the value of $$x$$ (or $$y$$) the get the single numerical value of $$2x^3$$. So statement (1) is not sufficient.

(2) y=0 --> clearly insufficient.

(1)+(2) $$y=0$$, so $$x=0$$ too (as $$x^6-y^6=0$$) and $$x^3-y^3=0$$. Sufficient.

Hope it's clear.
_________________
Manager
Joined: 27 Jul 2010
Posts: 197
Location: Prague
Schools: University of Economics Prague
Followers: 1

Kudos [?]: 24 [0], given: 15

Re: Powers doubt [#permalink]  24 Dec 2010, 03:46
This is not tough question, but you need to be carefull not to fall into trap of ODD POWER of numbers. you always need to be suspicious when you see one.
_________________

You want somethin', go get it. Period!

Re: Powers doubt   [#permalink] 24 Dec 2010, 03:46
Similar topics Replies Last post
Similar
Topics:
what is (x^3 + y^3 + z^3) / xyz 1 16 May 2013, 06:31
6 If n and m are integers, and x=3^n, and y=3^m, is the value 6 12 Sep 2012, 18:09
3 What is the value of y^3 - t^3? 8 23 Nov 2011, 08:57
2 If x^3*y = 24, what is the value of x^2*y^3 - x^2*y^2? 12 20 Apr 2010, 06:27
If X^3 *Y = 24 what is the value of X^3 * Y^3 - X^2 * Y^2? 16 17 Apr 2008, 08:16
Display posts from previous: Sort by