16*. Among the numbers $a, b$, and $c$, one is positive, one is negative, and one is zero. Additionally, it is known that $|a|=b^{2}(b-c)$. Which of the numbers is positive, which is negative, and which is zero?
16. Solution. If $b=0$, then $a=0$, which contradicts the condition. If $a=0$, then either $b=a=0$ or $b=c$, which in both cases contradicts the condition. The only remaining case is $c=0$. Then the given equation takes the form: $|a|=b^{3}$. Since $|a|>0$, $b$ must be a positive number. Therefore, $a \neq 0, b>0, c=0$. Other approaches to solving the problem are also possible.