Problem Solving

Is A>O?
(1)A³-A< 0
(2)1-A²> 0

This problem is from the Manhattan Guides. Just a real quick clarification that I need. I've figured the answer is C. But I did it in a much simpler way as compared to the complicated explanation given in the book.

From (1)
A(A²-1)<0

Meaning either A>0 and (A²-1)<0
Or A<0 and (A²-1)>0
This gives us conflicting values of A and hence INSUFFICIENT. Is my reasoning sufficient? Do I need to do any further testing?

From (2)
A²<1
|A|<1
-1<A<1
Again giving us conflicting values hence INSUFFICIENT.

Combining both
(2) Can be written as -A²+1> 0
Multiplying by -1 we get (A²-1)<0
Using this piece of information in statement 1, Since (A²-1)<0 then A has to be positive making A>0
Hence C. I would appreciate a clarification on the bold highlighted portion. Thanks

From (1)
A(A²-1)<0

Meaning either A>0 and (A²-1)<0
Or A<0 and (A²-1)>0
This gives us conflicting values of A and hence INSUFFICIENT. Is my reasoning sufficient? Do I need to do any further testing?

Hi,
I don't clearly understand what you mean by conflicting values. Anyway, my take on this:
a>0 and a^2-1 <0 => 0<a<1
a<0 and a^2-1>0 => a<-1
So, from (1) we get a<-1 or 0<a<1.
So, a can be either positive or negative. If this is what you meant by conflicting values, then it is perfect.