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X^8 - Y^8 =

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X^8 - Y^8 =  [#permalink]

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New post 13 Jul 2010, 17:42
3
19
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

71% (00:47) correct 29% (00:54) wrong based on 754 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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Re: MGMAT Challenge Problem  [#permalink]

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New post 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:

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

So, the answer is B.
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Re: MGMAT Challenge Problem  [#permalink]

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New post 13 Jul 2010, 18:39
4
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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Re: MGMAT Challenge Problem  [#permalink]

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New post 14 Jul 2010, 04:44
Looks like 500-level question))
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Re: MGMAT Challenge Problem  [#permalink]

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New post 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! :wink:
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Re: MGMAT Challenge Problem  [#permalink]

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New post 15 Jul 2010, 23:24
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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New post 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
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Re: X^8 - Y^8 =  [#permalink]

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New post 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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Re: X^8 - Y^8 =  [#permalink]

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New post 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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Re: X^8 - Y^8 =  [#permalink]

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

\(x^8-y^8\)

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

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

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

Answer B.
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Re: X^8 - Y^8 = &nbs [#permalink] 16 Nov 2018, 13:47
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