a^2/b=-8 then what is the value of a(a-b)

Given: \(\frac{a^2}{b}=-8\) --> \(a^2=-8b\). Q: \(a(a-b)=?\)

(1) \(\frac{a}{b}=-2\) --> as per Kaplan: \(4b^2=-8b\). Usually you can not divide equation by variable (or expression with variable) if you are not certain that variable (or expression with variable) does not equal to zero. So generally we should write: \(4b^2=-8b\) --> \(b^2+2b=0\) --> \(b(b+2)=0\) --> \(b=0\) or \(b=-2\). But in this case we can exclude \(b=0\) as a solution, as given that \(\frac{a^2}{b}=-8\) (and also in the statement \(\frac{a}{b}=-2\)): here \(b\) is in denominator and it can not be zero as in this case it won't make any sense (division by zero is undefined). So that is why Kaplan is safely dividing by \(b\). Though I think it's not a good question as, almost always, when GMAT provides some fraction and there is variable in denominator it states that variable does not equal to zero. As for the sign of \(b\): for equation it does not matter, you should only know that it does not equal to zero. Sign matters when you are dividing an inequality: in this case you should not only know that the variable (or expression with variable) does not equal to zero but also the sign of this variable (or expression with variable) to do the safe division/multiplication.

Now in our original question as only valid value of \(b\) is \(b=-2\) --> \(a=4\) and you can calculate \(a(a-b)\).

So this statement is sufficient.

(2) \(a-b=6\) --> \(a=b+6\) --> \(b^2+20b+36=0\) --> \(b=-18\) or \(b=-2\). Two values for \(b\), hence two values for \(a\) --> two values for \(a(a-b)\). Not sufficient.

Hope it's clear.