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x^3+y^3=?

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x^3+y^3=?  [#permalink]

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New post 26 Apr 2016, 06:36
1
3
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

79% (01:13) correct 21% (01:03) wrong based on 47 sessions

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x^3+y^3=?

1) x+y=0
2) x^2-y^2=0


* A solution will be posted in two days.

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Re: x^3+y^3=?  [#permalink]

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New post 26 Apr 2016, 07:00
MathRevolution wrote:
x^3+y^3=?

1) x+y=0
2) x^2-y^2=0


* A solution will be posted in two days.



always remember the rule \(x^n+y^n\)is never div by x-y..
It is div by x+y ONLY when n is ODD..


lets see the statement

1) x+y=0
As \(x^3+y^3\)has x+y as one of its factor , it is equal to 0..
suff

2)\(x^2-y^2=0\)..
\(x^2-y^2= (x-y)(x+y)=0\)..
we cannot say if x-y = 0 OR x+y = 0..
Insuff

A
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Re: x^3+y^3=?  [#permalink]

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New post 26 Apr 2016, 13:28
If you have trouble with the math on a problem like this, you can also do some trickery with the statements.

Notice that statement 1 already give us the info in statement 2. In other words, if x+y=0, then (x+y)(x-y) will have to equal 0 as well. That means that we can rule out B and C as answers. B is out because if statement 2 is sufficient, then statement 1, which gives the same info with more specifics (we know which term is 0), can't be insufficient. C is out because we can figure s2 out for ourselves from statement 1 alone. Therefore, there's no need to combine.

That leaves us with A, D, and E. So we're faced with a simpler decision. Can we solve from knowing that x^2 -y^2 = 0? If so, the answer is D. Do we need to know specifically that x+y=0? If so, then A is the answer. Is even that not enough? Then we go with E.

If you're curious, x^ + y^3 factors to (x+y)(x^2 -xy +y^2), so if we know x+y=0, then we know the whole thing is 0.

(Notice that the longer term is NOT equal to (x-y)^2, which has -2xy in the middle, not just -xy.)
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Re: x^3+y^3=?  [#permalink]

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New post 27 Apr 2016, 03:13
MathRevolution wrote:
x^3+y^3=?

1) x+y=0
2) x^2-y^2=0


* A solution will be posted in two days.

went with a^3 + b^3 = (a+b)(a^2-ab+b^2)
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Re: x^3+y^3=?  [#permalink]

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New post 29 Apr 2016, 20:10
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

x^3+y^3=?

1) x+y=0
2) x^2-y^2=0


-> In the original condition, there are 2 variables(x,y), which should match with the number of equations.
For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer.
When 1) & 2), you can only substitute 1)x=-y.
Then, x3^+y^3=(-y)^3+y^3=-y^3+y^3=0 is derived, which is unique and sufficient.
Hence, the answer is A.
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Re: x^3+y^3=?  [#permalink]

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New post 29 Apr 2016, 23:25
But an equal number of variables and independent equations does not guarantee a solution! We especially have to watch out for quadratics. In fact, we can't find the value of x or y here. We just know that one is the opposite of the other, and that's from one equation.

Or what about this?

x+y=5
xy=4

One of the variables is 4 and the other is 1, but which is which? The "one equation per variable" rule of thumb often leads us into traps on the GMAT, and the problem you've posted here is no exception.
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Re: x^3+y^3=?  [#permalink]

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New post 30 Apr 2016, 02:46
[quote="MathRevolution"]x^3+y^3=?

1) x+y=0
2) x^2-y^2=0


from 1st equation we know that a+b=0. cube both sides that means (a+b)^3=0
open the cube: a^3+b^3+3ab(a+b).
we know that a+b is 0 so 3ab(a+b) is 0
that means a^3+b^3=0
2nd information does give any solution.
so ans is A.
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Re: x^3+y^3=?  [#permalink]

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Re: x^3+y^3=?   [#permalink] 19 Apr 2019, 05:45
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