is x<0?

stmt1

x^3(1-x^2)<0

There are 3 cases a) x > 0 ( positive) then equation x^3(1-x^2)<0 holds true as (1-x^2) part of the equation will always be negative

b) x < 0 ( negative) then we are contradicting the equation x^3(1-x^2)<0 given as x^3 will be negative and 1-x^2) part of the equation will always be negative - So their multiplication will lead to a number greater than 0. ( Which is contradiction of the statement given)

Replace x with numbers ( integers, fractions ) it will reflect the results

According to me this statement ( Statement 1 ) is sufficient to deduce if x is negative or positive

Statement 2 x^2-1<0

then x^2<1 .. This case will be true only if 0<x<1 as for any other case square of a number is always positive. If x = 0 then statement holds true. So statement 2 is not sufficient to answer this question.

In my opinion , answer will be D.

I just started revisting my basics in maths after a long time and recently gearing up to start up my GMAT prep.

Would really appreciate your comments on it

Thanks

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