If xy ≠ 0, is x^3 + y^3 > 0 ?

(1) x + y > 0
(2) xy > 0

NEW question from GMAT® Official Guide 2019

(DS15561)
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1) Suff, if the sum is >0 , sum of cubes will also be >0 , we can test cases , only way for sum to be positive is by having greater magnitude of positive number ,hence the sum of cube will be more positive than negative cube , hence always be positive

2) Not suff, this tells that both the numbers have same sign, but for negative cases the sum of cube will be more negative , hence not suff

A
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Question-If xy ≠ 0, is x^3 + y^3 > 0 ?- can be further drill down :- In order to x^3 + y^3 > 0, Is X, and Y - Both are positive, or X is positive, and greater than Y, if Y is negative


1) x + y > 0
With the help of this statement we can say that either Both X, and Y are Positive, or X is positive, and greater than Y, if Y is negative
Sufficient

2) xy>0
Means both either positive, or Negative.

Hence, Insufficient

If xy ≠ 0, is x^3 + y^3 > 0 ?

(1) x + y > 0
(2) xy > 0

Statement 1 is insuff. as x and y can be either both positive number or one positive and other negative.
Statemtn 2 just tell us that the sign of both varibale are same, hence insuff

combining both we get to know that both x and y has same sign hence the answer is C.

updating- realized my mistake, A is correct.
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Answer should be A. If x+y>0, then atleast one number is poistive and the magnitude of this number will be greater than the magnitude of the other number. Hence the cube of this number will also be greater than teh cube of the other number. Hence x^3+Y^3>0. Sufficient

here is my reasoning

(1) x + y > 0

if x + y > 0 take numbers 2: -1 this implies that X must be bigger and positive, in case Y is negative Answer is Always YES. Sufficient.


(2) xy > 0

here both numbers can be negative say X is -4 and y is -2 anserr is NO , if we take X is 4; and Y is 2 answer is YES. Insufficient

Hi,

For someone like me who is not good at number picking ( i must improve this as it costs me a lot), here is alebric approach

We know that
\((x+y)^3= x^3+y^3+3xy(x+y)\)
or
\(x^3+y^3= (x+y)^3- 3xy(x+y)\)

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

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

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

so if we get to know that either (x+y) & \((x^2+y^2 -xy)\)is >0 or both are <0 then we can conclude about \(x^3+y^3\)

Now statement A :
x+y >0

this means
x>0, y>0 so xy>0 hence we can see that (x+y) >0 .
Ok what about this exp \((x^2+y^2 -xy)\) . For x>0 and y>0 we will have x^2>0, y^2>0 and xy>0 but we can always see that x^2+y^2> xy
so we have \((x^2+y^2 -xy)\) also >0 Hence we can conclude that \(x^3+y^3\)>0
or
x>0, y<0 and |x|>|y| , so xy would be <0 , so we can see that (x+y) >0 & \((x^2+y^2 -xy)\) >0 hence we can say that \(x^3+y^3\)>0

or
x<0, y>0 and |y|>|x| , so xy would be <0 , so we can see that (x+y) >0 & \((x^2+y^2 -xy)\) >0 hence we can say that \(x^3+y^3\)>0

Hence A is sufficient.

Now Statement B:
its says xy>0
all i can infer is either x>0&y>0 or x<0&y<0

but for this (x+y) can be either >0 or the exp (x+y) <0 So really cant conclude that the exp \(x^3+y^3\)>0

We have already seen that if x+y>0 then we can conclude that \(x^3+y^3\)>0

But if x+y<0 , then the exp \((x^2+y^2 -xy)\) must also be <0 for\(x^3+y^3\)>0
But we can have that \(|x^2+y^2|\)<|xy|. in that case we will have \(x^3+y^3\)>0
or\(|x^2+y^2|\)>|xy| in that case we will have \(x^3+y^3\)<0

hence B is insufficient.

Probus
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