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Re: X^8 - Y^8 = [#permalink]
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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. \((X^2 - Y^2)^4\)


Since \((X^4)^2 = X^8\), and \((Y^4)^2 = Y^8\), we can see that \(X^8 - Y^8\) is a difference of squares

So, \(X^8 - Y^8 =(X^4 + Y^4)(X^4 - Y^4)\)
Scan the answer choices....not there!!

Aha, now we must recognize that \(X^4 - Y^4\) is a difference of squares, which we can factor to get:
\((X^4 + Y^4)(X^4 - Y^4) = (X^4 + Y^4)(X^2 + Y^2)(X^2 - Y^2)\)

Since \(X^2 - Y^2\) is another difference of squares, we can continue factoring to get...
\((X^4 + Y^4)(X^2 + Y^2)(X^2 - Y^2)=(X^4 + Y^4)(X^2 + Y^2)(X+Y)(X-Y)\)

Answer: B
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Re: MGMAT Challenge Problem [#permalink]
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Looks like 500-level question))
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Re: MGMAT Challenge Problem [#permalink]
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]
I go with option B

It is definitely not a 700 level problem
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Re: X^8 - Y^8 = [#permalink]
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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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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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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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 = [#permalink]
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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. \((X^2 - Y^2)^4\)


X^8 - Y^8 =

(X^4 + Y^4)(X^4 - Y^4)

Notice that the second expression is a difference of fourth powers, so it can be factored further as:

(X^4 + Y^4)(X^2 - Y^2)(X^2 + Y^2)

The middle expression is a difference of squares, so it can be factored further as:
(X^4 + Y^4)(X - Y)(X + Y)(X^2 + Y^2)

Answer: B
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Re: X^8 - Y^8 = [#permalink]
\(X^8 - Y^8 = (X^4+Y^4)(X^4-Y^4) = (X^4+Y^4)(X^2+Y^2)(X^2-Y^2) = \\
(X^4+Y^4)(X^2+Y^2)(X+Y)(X-Y)\)

IMO B

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