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# If r^2 - 2rs+ s^2 = 4, then (r - s)^6 =

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Math Expert
Joined: 02 Sep 2009
Posts: 58453
If r^2 - 2rs+ s^2 = 4, then (r - s)^6 =  [#permalink]

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10 Apr 2019, 21:43
00:00

Difficulty:

5% (low)

Question Stats:

96% (00:58) correct 4% (01:03) wrong based on 57 sessions

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If $$r^2 - 2rs+ s^2 = 4$$, then $$(r - s)^6 =$$

(A) -4
(B) 4
(C) 8
(D) 16
(E) 64

Source: Nova GMAT
Difficulty Level: 500

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Joined: 18 Aug 2017
Posts: 5020
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: If r^2 - 2rs+ s^2 = 4, then (r - s)^6 =  [#permalink]

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11 Apr 2019, 00:43
1
Bunuel wrote:
If $$r^2 - 2rs+ s^2 = 4$$, then $$(r - s)^6 =$$

(A) -4
(B) 4
(C) 8
(D) 16
(E) 64

$$r^2 - 2rs+ s^2 = 4$$

we get (r-s)^2 = 4
or say r-s = +/-2
so (r-s)^6 = 64
IMO E
Intern
Joined: 26 May 2019
Posts: 7
Re: If r^2 - 2rs+ s^2 = 4, then (r - s)^6 =  [#permalink]

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15 Jun 2019, 01:17
As (r−s)6

64 is the only option which can be written in the form 2^6
Re: If r^2 - 2rs+ s^2 = 4, then (r - s)^6 =   [#permalink] 15 Jun 2019, 01:17
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