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4(xy)^3 + (x^3 − y^3)^2 =

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Bunuel
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4(xy)^3 + (x^3 − y^3)^2 = [#permalink] New post   Updated on: 08 Jul 2019, 05:35
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\(4(xy)^3 + (x^3 − y^3)^2 =\)


(A) \(x^3 − y^3\)

(B) \((x^2 + y^2)^3\)

(C) \((x^3 + y^3)^3\)

(D) \((x^3 − y^3)^2\)

(E) \((x^3 + y^3)^2\)

Source: Nova GMAT
Difficulty Level: 550
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E

Originally posted by Bunuel on 01 Jun 2017, 12:00.
Last edited by Sajjad1994 on 08 Jul 2019, 05:35, edited 1 time in total.
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generis
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Re: 4(xy)^3 + (x^3 − y^3)^2 = [#permalink] New post   01 Jun 2017, 12:52
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Bunuel wrote:
\(4(xy)^3 + (x^3 − y^3)^2 =\)


(A) \(x^3 − y^3\)

(B) \((x^2 + y^2)^3\)

(C) \((x^3 + y^3)^3\)

(D) \((x^3 − y^3)^2\)

(E) \((x^3 + y^3)^2\)


\(4(xy)^3 + (x^3 − y^3)^2 =\) - expand the Square of a Difference:

\(4(xy)^3\) + \(x^6 - (2)(xy)^3 + y^6\) =

\(x^6 + (2)(xy)^3 + y^6\) = the expanded version of Square of a Sum ==>

\((x^3 + y^3)^2\)

Answer
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E
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Re: 4(xy)^3 + (x^3 − y^3)^2 = [#permalink] New post   04 Jun 2017, 12:44
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We have

\(4(xy)^3 + (x^3 - y^3)^2 = > 4(xy)^3 + x^6 - 2(xy)^3 + y^6 = x^6 + 2(xy)^3 + y^6 = (x^3+y^3)^2\)

Therefore, the answer is option E

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Re: 4(xy)^3 + (x^3 − y^3)^2 = [#permalink] New post   05 Jun 2017, 00:42
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This can also be solved by plugging in values. Lets assume x = y = 1.
So 4(xy)^3 + (x^3 − y^3)^2 = 4 *(1*1)^2 + (1-1)^2 = 4.

Thus we have to look for an option where when we put x=1 and y=1, we should get our answer as '4'.

Check A) 1-1 = 0
B) (1+1)^3 = 8
C) (1+1)^3 = 8
D) (1-1)^2 = 0
E) (1+1)^2 = 4

Only E satisfies. Thus E answer
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Re: 4(xy)^3 + (x^3 − y^3)^2 = [#permalink] New post   05 Jun 2017, 03:18
amanvermagmat wrote:
This can also be solved by plugging in values. Lets assume x = y = 1.
So 4(xy)^3 + (x^3 − y^3)^2 = 4 *(1*1)^2 + (1-1)^2 = 4.

Thus we have to look for an option where when we put x=1 and y=1, we should get our answer as '4'.

Check A) 1-1 = 0
B) (1+1)^3 = 8
C) (1+1)^3 = 8
D) (1-1)^2 = 0
E) (1+1)^2 = 4

Only E satisfies. Thus E answer


4x^3*y^3 + x^6 - 2(x^3- y^3) + y^6. On opening up the brackets we get, 4x^3*y^3 + x^6 - 2x^3y^3 + y^6. Cancel out 4x^3y^3 and 2x^3y^3. Finally we get, x^6 + 2x^3y^3 +y^6. Therefore, the answer is E.
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Re: 4(xy)^3 + (x^3 − y^3)^2 = [#permalink] New post   07 Jun 2017, 04:40
4(xy)^3 + (x^3-y^3)^2
= 4(xy)^3 + x^6 + y^6 -2(xy)^3
= x^6 + y^6 +2(xy)^3
= (x^3 + y^3)^2.

Option E is should be the correct answer, waiting for OA :)
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Re: 4(xy)^3 + (x^3 − y^3)^2 = [#permalink] New post   07 Jun 2017, 05:21
=4(xy)^3 + (x^6 - 2(xy)^3 + y^6)
=2(xy)^3 +x^6 +y^6
=(x^3 + y^3)^2

Hence, the answer should be E.

I would appreciate a kudos if you liked my solution!
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Re: 4(xy)^3 + (x^3 − y^3)^2 = [#permalink] New post   07 Jun 2017, 05:27
Let x=y=1

The expression \(4(xy)^3 + (x^3 − y^3)^2\) gives us a value of 4

Evaluating the answer choices, Only \((x^3+y^3)^2\)(Option E) gives us the same value
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Re: 4(xy)^3 + (x^3 − y^3)^2 = [#permalink] New post   28 Aug 2017, 00:21
4(xy)^3+x^6-2x^3y^3+y^3
=x^6+y^6+2x^3y^3
=(x^3+y^3)^2

Hence Option E
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Re: 4(xy)^3 + (x^3 − y^3)^2 = [#permalink] New post   31 Aug 2017, 10:21
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Bunuel wrote:
\(4(xy)^3 + (x^3 − y^3)^2 =\)


(A) \(x^3 − y^3\)

(B) \((x^2 + y^2)^3\)

(C) \((x^3 + y^3)^3\)

(D) \((x^3 − y^3)^2\)

(E) \((x^3 + y^3)^2\)


4(xy)^3 + (x^3 - y^3)^2

4x^3y^3 + x^6 - 2x^3y^3 + y^6

x^6 + 2x^3y^3 + y^6

(x^3 + y^3)^2

Answer: E
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Re: 4(xy)^3 + (x^3 − y^3)^2 = [#permalink] New post   21 Sep 2018, 12:54
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Re: 4(xy)^3 + (x^3 − y^3)^2 = [#permalink]
21 Sep 2018, 12:54
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