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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Probus

~You Just Can't beat the person who never gives up~ Babe Ruth