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(x^8y^2+x^2y^8)/(x^2+y^2)=?

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(x^8y^2+x^2y^8)/(x^2+y^2)=?  [#permalink]

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New post 31 Dec 2018, 02:25
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A
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Difficulty:

  55% (hard)

Question Stats:

67% (02:02) correct 33% (02:38) wrong based on 54 sessions

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[Math Revolution GMAT math practice question]

\(\frac{(x^8y^2+x^2y^8)}{(x^2+y^2)}=?\)

\(A. x^2y^2(x^4-x^2y^2+y^4)\)
\(B. x^2y(x^4+x^2y^2+y^4)\)
\(C. x^2y^2(x^2+y^2)\)
\(D. x^2y^2(x+y)\)
\(E. x^4y^4(x^2-y^2)\)

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Re: (x^8y^2+x^2y^8)/(x^2+y^2)=?  [#permalink]

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New post 31 Dec 2018, 04:59
1
(x^8y^2 + x^2y^8)/(x^2 + y^2) = ?

Step 1: Take x^2y^2 common from both the terms of the numerator
(x^2y^2)(x^6 + y^6)/(x^2 + y^2)

Step 2: (x^6 + y^6) can be represented as {(x^2)^3 + (y^2)^3}, i.e. in the form a^3 + b^3
(x^2y^2){(x^2)^3 + (y^2)^3}/(x^2 + y^2)

Step 3: Split a^3 + b^3 in to into factor formula, i.e. a^3 + b^3 = (a + b)(a^2 - ab + b^2)
(x^2y^2){(x^2 + y^2)(x^4 - x^2y^2 + y^4}/(x^2 + y^2)

Step 4: Cancel out (x^2 + y^2) from numerator and denominator
(x^2y^2)(x^4 - x^2y^2 + y^4)

Hence, the Correct Answer is Option A. (x^2y^2)(x^4 - x^2y^2 + y^4)

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Re: (x^8y^2+x^2y^8)/(x^2+y^2)=?  [#permalink]

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New post 31 Dec 2018, 05:22
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Assume x = 1 and y = 1 ===> Prompt = 2/2 = 1

Test x = 1 and y = 1 in the answers and only option A = 1. The correct answer is A.

MathRevolution wrote:
[Math Revolution GMAT math practice question]

\(\frac{(x^8y^2+x^2y^8)}{(x^2+y^2)}=?\)

\(A. x^2y^2(x^4-x^2y^2+y^4)\)
\(B. x^2y(x^4+x^2y^2+y^4)\)
\(C. x^2y^2(x^2+y^2)\)
\(D. x^2y^2(x+y)\)
\(E. x^4y^4(x^2-y^2)\)
Math Revolution GMAT Instructor
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Joined: 16 Aug 2015
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GMAT 1: 760 Q51 V42
GPA: 3.82
Re: (x^8y^2+x^2y^8)/(x^2+y^2)=?  [#permalink]

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New post 02 Jan 2019, 01:46
=>

\(\frac{(x^8y^2+x^2y^8)}{(x^2+y^2)}= \frac{x^2y^2(x^6+y^6)}{(x^2+y^2)}= \frac{x^2y^2(x^2+y^2)(x^4-x^2y^2+y^4)}{(x^2+y^2)}= x^2y^2 (x^4-x^2y^2+y^4)\), using the identity \(a^3+b^3 = (a+b)(a^2-ab+b^2)\).

Therefore, the answer is A.
Answer: A
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Re: (x^8y^2+x^2y^8)/(x^2+y^2)=?   [#permalink] 02 Jan 2019, 01:46
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