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# X^8 - Y^8 =

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SVP
Affiliations: HEC
Joined: 28 Sep 2009
Posts: 1636
Concentration: Economics, Finance
GMAT 1: 730 Q48 V44
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Kudos [?]: 647 [1] , given: 432

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13 Jul 2010, 18:42
1
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8
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Difficulty:

25% (medium)

Question Stats:

68% (01:46) correct 32% (00:54) wrong based on 576 sessions

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X^8 - Y^8 =

A. $$(X^4 - Y^4)^2$$
B. $$(X^4 + Y^4)(X^2 + Y^2)(X + Y)(X - Y)$$
C. $$(X^6 + Y^2)(X^2 - Y^6)$$
D. $$(X^4 - Y^4)(X^2 - Y^2)(X - Y)(X + Y)$$
E. $$(X^2 - Y^2)^4$$
[Reveal] Spoiler: OA

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Senior Manager
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13 Jul 2010, 19:39
4
KUDOS
Apply a^2 - b^2 = (a-b)(a+b)

(x^4)^2 - (y^4)^2

( x^4 + y^4) ( x^4 - y^4)

( x^4 + y^4) (x^2+y^2)(x^2-y^2)

( x^4 + y^4) (x^2+y^2)(x+y)(x-y)

So B
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Concentration: Marketing, General Management
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14 Jul 2010, 05:44
Looks like 500-level question))
Ms. Big Fat Panda
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14 Jul 2010, 06:13
2
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3
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The solution is just a basic application of the principle that $$x^2 - y^2 = (x+y)(x-y)$$

So, when we simplify, we treat $$x^8 = (x^4)^2 = ((x^2)^2)^2$$

So we get the following:

$$x^8 - y^8 = (x^4+y^4)(x^4-y^4) = (x^4+y^4)(x^2+y^2)(x^2 - y^2) =(x^4+y^4)(x^2+y^2)(x+y)(x-y)$$

SVP
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Concentration: Economics, Finance
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Kudos [?]: 647 [0], given: 432

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14 Jul 2010, 06:56
ulm wrote:
Looks like 500-level question))

Agreed. I personally think that it belongs in the 600-level category, but MGMAT considers it a 700-level question. I just wanted to stay as faithful to the question source as possible.

I'll post a harder one to make up for it!
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16 Jul 2010, 00:24
I go with option B

It is definitely not a 700 level problem
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Re: X^8 - Y^8 = [#permalink]

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24 Aug 2014, 17:04
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: X^8 - Y^8 = [#permalink]

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24 Aug 2014, 23:31
BM wrote:
X^8 - Y^8 =

A. $$(X^4 - Y^4)^2$$
B. $$(X^4 + Y^4)(X^2 + Y^2)(X + Y)(X - Y)$$
C. $$(X^6 + Y^2)(X^2 - Y^6)$$
D. $$(X^4 - Y^4)(X^2 - Y^2)(X - Y)(X + Y)$$
E. $$(X^2 - Y^2)^4$$

I solved this is hardly 3 seconds, would tell how?

$$x^8 - y^8$$

It has even power($$8 = 2^3$$) & -ve sign in between, so fully expandable to the least power of x & y i.e (x+y)

All +ve powers cannot be expanded, so out of the 5 options, just search for (x-y) & terms with high +ve powers

Only option B fits in
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Re: X^8 - Y^8 = [#permalink]

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01 Sep 2015, 05:15
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: X^8 - Y^8 = [#permalink]

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06 Sep 2016, 08:01
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: X^8 - Y^8 =   [#permalink] 06 Sep 2016, 08:01
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