Hello
Please note b2 is square of b,
I have analyzed the following question & the answer to it.
However, I fail to understand why square of b is assumed to be positive. It can be zero also.
Thus answer should be E.
Please explain this.

thanks in advance.
Hamps
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Is a > c?

(1) b > d

(2) ab2 – b > b2c – d
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Recommended solution:
This is a multiple variable inequality problem, so you must solve it by doing algebraic manipulations on the inequalities.

(1) INSUFFICIENT: Statement (1) relates b to d, while giving us no knowledge about a and c. Therefore statement (1) is insufficient.

(2) INSUFFICIENT: Statement (2) does give a relationship between a and c, but it still depends on the values of b and d. One way to see this clearly is by realizing that only the right side of the equation contains the variable d. Perhaps ab2 – b is greater than b2c – d simply because of the magnitude of d. Therefore there is no way to draw any conclusions about the relationship between a and c.

(1) AND (2) SUFFICIENT: By adding the two inequalities from statements (1) and (2) together, we can come to the conclusion that a > c. Two inequalities can always be added together as long as the direction of the inequality signs is the same:

ab2 – b > b2c – d
(+) b > d
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ab2 > b2c

Now divide both sides by b2. Since b2 is always positive, you don't have to worry about reversing the direction of the inequality. The final result: a > c.

The correct answer is C.
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it's good that you're aware of exceptional cases like this, but, unfortunately, there is no such exceptional case in this problem.
you see, if b = 0, then a(b^2) can't be greater than (b^2)c, as stated in the inequality above; both sides would just be zero, and would therefore be equal.

since the left side is greater than the right side, they can't both be zero; therefore, b is not zero.

however, it is excellent that you are considering the possibility that b = 0 in the first place; many a test taker is broadsided by such exceptional cases, especially on data sufficiency problems (which can be turned completely upside down by just one such instance).