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(1) \(x^3 \lt y^3\). Note that we can always take an odd root from both sides of an inequality (the same for raising both parts of an inequality to an odd power). Now, if we take 3rd root from \(x^3 \lt y^3\) we'll get \(x \lt y\). Sufficient.

(2) \((x+y)(x-y) \lt 0\). This statement tells that \(x^2-y^2 \lt 0\) or \(x^2 \lt y^2\), which is not sufficient to answer whether \(x \lt y\), consider \(x=1\) and \(y=2\) for an YES answer and \(x=1\) and \(y=-2\) for a NO answer.

(1) \(x^3 \lt y^3\). Note that we can always take an odd root from both sides of an inequality (the same for raising both parts of an inequality to an odd power). Now, if we take 3rd root from \(x^3 \lt y^3\) we'll get \(x \lt y\). Sufficient.

(2) \((x+y)(x-y) \lt 0\). This statement tells that \(x^2-y^2 \lt 0\) or \(x^2 \lt y^2\), which is not sufficient to answer whether \(x \lt y\), consider \(x=1\) and \(y=2\) for an YES answer and \(x=1\) and \(y=-2\) for a NO answer.

Answer: A

Hi, why can we not consider that either (x+y)<0 or (x-y)<0? If we do this then we get X<-Y and X<Y. Thx

(1) \(x^3 \lt y^3\). Note that we can always take an odd root from both sides of an inequality (the same for raising both parts of an inequality to an odd power). Now, if we take 3rd root from \(x^3 \lt y^3\) we'll get \(x \lt y\). Sufficient.

(2) \((x+y)(x-y) \lt 0\). This statement tells that \(x^2-y^2 \lt 0\) or \(x^2 \lt y^2\), which is not sufficient to answer whether \(x \lt y\), consider \(x=1\) and \(y=2\) for an YES answer and \(x=1\) and \(y=-2\) for a NO answer.

Answer: A

Hi Bunuel,

For 2, why can't I write -y<x<y ? Is it because we don't really know the sign of y?

(1) \(x^3 \lt y^3\). Note that we can always take an odd root from both sides of an inequality (the same for raising both parts of an inequality to an odd power). Now, if we take 3rd root from \(x^3 \lt y^3\) we'll get \(x \lt y\). Sufficient.

(2) \((x+y)(x-y) \lt 0\). This statement tells that \(x^2-y^2 \lt 0\) or \(x^2 \lt y^2\), which is not sufficient to answer whether \(x \lt y\), consider \(x=1\) and \(y=2\) for an YES answer and \(x=1\) and \(y=-2\) for a NO answer.

Answer: A

Hi Bunuel,

For 2, why can't I write -y<x<y ? Is it because we don't really know the sign of y?

Regards

\(x^2 \lt y^2\) means that |x| < |y|, so y is further from 0 than x is. We can have the following cases: