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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From the question \(x^3+y^3>0\). I think of the possible cases that will make the equation is true which are
1. X and Y are positive value
2. X is positive value with higher magnitude than Y
e.g. 2 + (-1) > 0

(1) x + y > 0
Let us know that x+y is positive value which let us know the possible case of x and y are
1. X and Y are positive value
2. X is positive value with higher magnitude than Y
which is the same as our previous analysis ---> Sufficient

(2) xy > 0
This statement show us that xy is positive value which let us know the possible case of x and y are
1. X and Y are positive value --> e.g. 2*3>0
2. X and Y are negative value --> e.g. -2*-3>0
This statement is not sufficient since \(x^3+y^3\) can result in both positive or negative values ---> insufficient

So my answer is "A"

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

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

= \((x+y)[x^2+\frac{(y^2)}{4}-xy+\frac{3(y^2)}{4}]\)

= \((x+y)[(x-\frac{y}{2})^2}+\frac{3(y^2)}{4}]\)
as\([(x-\frac{y}{2})^2}+\frac{3(y^2)}{4}]\)will always be positive as x,y≠0
Hence \(x^3+y^3\) will take the same sign as (x+y)
Option A is correct

A for me, because it is sufficient to know that x + y > 0 to draw the conclusion that x^3+y^3>0

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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Statement I is sufficient to answer the equation.

I agree with your solution

OH MY LORD

I was trying to ask whether the equation is more than 1 or not, which was not what asked in the question.

My blunder....I do apologize.

the question got me and I couldn't see my blunder and made 2 posts...I think I should take some rest...HAHAAHA