If xy 0, is x^3 + y^3 > 0 ?

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

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

I did it slightly different than the methods above for option A (correct ans)

x^3 + y^3 >0 ---> x^3 > -y^3

if we can arrange A in the same pattern and cube it, we get x^3 > -y^3

B ans yes/ no as x,y can be +ve or -ve

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

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

Statement 1: \(x + y > 0\)

This is possible if

#1. Both \(x\) and \(y\) are positive => \(x^3 \)and\( y^3\) are positive \(=> x^3 + y^3 > 0\)

#2. One of \(x\) and \(y\) is negative (say \(y\)) and the other is positive (say \(x\)), with the positive number having a higher magnitude

=> \(x^3\) is positive and \(y^3\) is negative, but the magnitude of \(x^3\) is higher than that of \(y^3\)\(=> x^3 + y^3 > 0\)

Thus, statement 1 is sufficient

Statement 2: \(xy > 0\)

This is possible if:

#1. Both \(x\) and \(y\) are positive => \(x^3 \)and\( y^3\) are positive \(=> x^3 + y^3 > 0\)

#2. Both \(x\) and \(y\) are negative => \(x^3 \)and\( y^3\) are negative \(=> x^3 + y^3 < 0\)

Thus, statement 2 is not sufficient


Answer A

Target question: Is x³ + y³ > 0?

Statement 1: x + y > 0
There are two possible cases to consider here:
case i: x and y are both positive
case ii: One value (x or y) is positive and the other value is negative, AND the magnitude (i.e., absolute value) of the positive value is greater than the magnitude of the negative value

If case i is true, then it is clear that x³ + y³ well be greater than 0.
What about case ii?

Here's an important property: If |a| < |b|, then |a²| < |b²| and |a³| < |b³| and |a⁴| < |b⁴|, . . . . etc

Let's make it easy on ourselves and say x is the positive number and y is the negative number.
If x + y > 0, then we know that |y| < |x|
From the above property, we can conclude that |y³| < |x³|
If the magnitude of x³ (which is positive) is greater than the magnitude of y³ (which is negative), we can be certain that x³ + y³ > 0
In other words, the answer to the target question is YES, x³ + y³ is greater than 0
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: xy > 0
There are several values of x and y that satisfy statement 2. Here are two:
Case a: x = 1 and y = 1. In this case, the answer to the target question is YES, x³ + y³ is greater than 0
Case b: x = -1 and y = -1. In this case, the answer to the target question is NO, x³ + y³ is not greater than 0
Since we can’t answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent

Bunuel

for this, is it possible to simplify the equation?

x^3 > -y^3
take the cube root
x > -y

1. x+y> 0
x > -y
Sufficient

2. xy > 0
x pos, y pos = OK
x neg, y neg = No
Not sufficient . A

Posted from my mobile device

Yes, that's a correct method. +1.

Video solution from Quant Reasoning:
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Solution:

Question Stem Analysis:

We need to determine whether x^3 + y^3 > 0, given that neither x nor y is 0.

Statement One Only:

Since x + y > 0, either both x and y are positive or one of them is positive and the other is negative. If both x and y are positive, then obviously x^3 + y^3 > 0. If one of them is positive and the other is negative, without loss of generality, we can let x be positive and y be negative. In order for x + y > 0, x must be “more” positive than y is negative (in other words, |x| > |y|) . In that case, x^3 will also be “more” positive than y^3 (in other words, |x^3| > |y^3|). Therefore, x^3 + y^3 > 0. Statement one alone is sufficient.

(Alternatively, recall that x^3 + y^3 = (x + y)(x^2 - xy + y^2). Again, if both x and y are positive, then obviously x^3 + y^3 > 0. Now, if one of the values is positive and the other is negative, then xy is negative and hence -xy is positive. Since x^2 + y^2 is positive, x^2 - xy + y^2 will be positive. Therefore, given the fact that x + y is positive, we see that (x + y)(x^2 - xy + y^2) will be positive, i.e., x^3 + y^3 is positive.)

Statement Two Only:

Since xy > 0, either both x and y are positive or both are negative. If both x and y are positive, then obviously x^3 + y^3 > 0. However, if both are negative, then x^3 + y^3 < 0. Statement two is not sufficient.

Answer: A

Answer: Option A

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Let's work on question stem first: xˆ3+yˆ3 = (x+y)(xˆ2+yˆ2- xy)
Question asks: (x+y)(xˆ2+yˆ2-xy)>0, it means are (x+y) and (xˆ2+yˆ2- xy) of same sign or different.

S1: x+y >0,
Case I: both positive, which gives us (xˆ2 + yˆ2 - xy) positive. Try with any value, it will be positive because we have a summation of the square of two number & their multiplication.
Case II: one positive one negative. In this case also (xˆ2 + yˆ2 - xy) will be positive because square of the numbers will be positive and -xy will be positive.

Both cases satisfy S1, hence sufficient.



S2: xy>0

Case I: if x & y both are negative, (x+y) will be negative & (xˆ2 + yˆ2 - xy) will be positive (as per Case I, S1). Hence, (x+y)(xˆ2+yˆ2-xy) <0
Case II: if x& y both positive, then (x+y)(xˆ2+yˆ2-xy)>0, (Check Case S1).
solution for Case I & II is different. Hence S2 is not sufficient.

Mark A as a answer

Hi, can you validate if my approach is correct?

Pre-solving- x^3 + y^3 > 0

Subtract y^3 from both sides -> x3 > -y^3

Take cube root -> Is x > -y?


St1- x+y > 0

x>-y.... Sufficient.

St2- xy >0

Either x & y are both positive or both negative.

Here I took examples and replaced them back in the original q stem -> x^3 + y^3 > 0


-2*-3>0
6>0

So; -8 + (-27) which is not greater than 0

Or 2*3>0
6>0

So; 8 + 27 which is greater than 0

Hence S2 is not suff.

Ans A.

Please validate chetan2u Bunuel

Hi
You are correct in your approach.

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