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