It's a problem on inequality. You cannot expand the expression and equate it to zero!

1: Square of any positive or negative real number is always positive. Thus, (x-3) and x can be both positive or negative.

1) x-3>0 => x>3 => x is positive

2) x-3<0 => x<3 => x can have negative values too!

(x-3)^2 > 0 cannot imply that x is positive and thus is not sufficient.

2: x^3-1> => x^3>1. Cube of a positive number is always positive, and of a negative number is always negative. This is sufficient, as it gives us x to be positive.

Thus the correct answer is B, statement 2 is sufficient.

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