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If x^2 + y^2 = 16 - 2xy, then (x + y)^4 =

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If x^2 + y^2 = 16 - 2xy, then (x + y)^4 = [#permalink] New post 24 Mar 2018, 05:11
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If \(x^2 + y^2 = 16 - 2xy\), then \((x + y)^4 =\)

A. 4

B. 32

C. 48

D. 64

E. 256
[Reveal] Spoiler: OA

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Re: If x^2 + y^2 = 16 - 2xy, then (x + y)^4 = [#permalink] New post 25 Mar 2018, 19:45
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Key to this one is noticing the hidden special equation. x^2 + y^2 isn't one, but it's close. If you shuffle the equation around a bit you get:

x^2 + 2xy + y^2 = 16

which can be reformatted to:

(x + y)^2 = 16

Since (x + y)^4 is (x + y)^2 squared, we can say that:

(x + y)^4 = 16^2

which is 256, giving us answer choice E.
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Re: If x^2 + y^2 = 16 - 2xy, then (x + y)^4 =   [#permalink] 25 Mar 2018, 19:45
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If x^2 + y^2 = 16 - 2xy, then (x + y)^4 =

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