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# X^8 - Y^8 =

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SVP
Affiliations: HEC
Joined: 28 Sep 2009
Posts: 1544
Concentration: Economics, Finance
GMAT 1: 730 Q48 V44

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13 Jul 2010, 17:42
4
23
00:00

Difficulty:

25% (medium)

Question Stats:

73% (01:08) correct 27% (01:27) wrong based on 567 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. $$(X^2 - Y^2)^4$$

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Joined: 08 Jan 2009
Posts: 205

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13 Jul 2010, 18:39
5
1
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
##### General Discussion
Manager
Joined: 03 Jun 2010
Posts: 120
Location: United States (MI)
Concentration: Marketing, General Management

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14 Jul 2010, 04:44
Looks like 500-level question))
SVP
Status: Three Down.
Joined: 09 Jun 2010
Posts: 1812
Concentration: General Management, Nonprofit

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14 Jul 2010, 05:13
2
3
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
Affiliations: HEC
Joined: 28 Sep 2009
Posts: 1544
Concentration: Economics, Finance
GMAT 1: 730 Q48 V44

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14 Jul 2010, 05:56
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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Joined: 12 Jun 2007
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15 Jul 2010, 23:24
I go with option B

It is definitely not a 700 level problem
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Joined: 27 Dec 2012
Posts: 1712
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: X^8 - Y^8 =  [#permalink]

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24 Aug 2014, 22:31
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
Intern
Joined: 02 Nov 2017
Posts: 17
Re: X^8 - Y^8 =  [#permalink]

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31 Mar 2018, 01:33
1
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. $$(X^2 - Y^2)^4$$

Bunuel could you please explain this?
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Status: Stepping into my 10 years long dream
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Re: X^8 - Y^8 =  [#permalink]

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31 Mar 2018, 04:24
shivamtibrewala wrote:
Bunuel could you please explain this?

Hey shivamtibrewala ,

It is the simple use of formula $$a^2 - b^2 = (a+b)(a-b)$$

Consider $$a = x^4$$ and $$b = y^4$$ first and expand as mentioned above.

Once done you will get $$(x^4 +y^4)(x^4 - y^4)$$

Now, this is again the usage of same formula. Here consider $$a = x^2$$ and $$b = y^2$$

Once done you will get $$(x^2+y^2)(x^2 - y^2)$$

Same goes for $$(x^2 - y^2)$$.

hence you will get :

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

Does that make sense?
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Intern
Joined: 29 Sep 2018
Posts: 11
Re: X^8 - Y^8 =  [#permalink]

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16 Nov 2018, 13:47
This question can be solved by taking the difference of squares multiple times.

$$x^8-y^8$$

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

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

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

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Re: X^8 - Y^8 =  [#permalink]

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08 Apr 2020, 14:21
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Re: X^8 - Y^8 =   [#permalink] 08 Apr 2020, 14:21
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