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![What is the value of (x^8y^2 + x^2y^8)/(x^2 + y^2)?]](/forum/styles/gmatclub_light/imageset/icon_post_target_unread.svg "What is the value of (x^8y^2 + x^2y^8)/(x^2 + y^2)?]"){kind=icon}
10 Nov 2023, 11:13
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
61% (02:18) correct
39%
(02:33)
wrong
based on 170
sessions
History
Date
Time
Result
Not Attempted Yet
What is the value of \(\frac{(x^8y^2 + x^2y^8)}{(x^2 + y^2)}\)?
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. \(x^4y^4(x^2-y^2)\)
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. \(x^4y^4(x^2-y^2)\)
10 Nov 2023, 23:10
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Bunuel
OA is A
Remember whenever these type of questions comes in question which involves algebra with higher power. Try to resolve using putting the value.
Three easy way you can use
1) Put x=y=1
Atleast 2 option will be eliminated for sure
2) Put x=1 and Y =0 or vice versa
If still not resolve
Then put x=1 y=2.
It will mostly resolve your question
Posted from my mobile device
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General Discussion
27 Jun 2025, 06:32
Kudos
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numerator = x^2 * y^2 (x^6 + y^6)
x^6 + y^6 can be factored again by sum of cubes
= (x^2 + y^2)(x^4 - x^2 y^2 + y^4)
now the (x^2 + y^2) cancels and you're left with
x^2 y^2 (x^4 - x^2 y^2 + y^4)
A
x^6 + y^6 can be factored again by sum of cubes
= (x^2 + y^2)(x^4 - x^2 y^2 + y^4)
now the (x^2 + y^2) cancels and you're left with
x^2 y^2 (x^4 - x^2 y^2 + y^4)
A
Bunuel
