Apostol Calculus, Method of Exhaustion In Apostol's Calculus, he goes through the method of exhaustion to find the area under a parabola from $0 \ to\  b$. Using the fact that,
\begin{align}
 &1^2+2^2+...+(n-1)^2 < \frac{n^3}{3} < 1^2+2^2+...+n^2\label{1} \\
 &\Rightarrow \frac{b^3}{n^3}(1^2+2^2+...+(n-1)^2) < \frac{b^3}{3} < \frac{b^3}{n^3}(1^2+2^2+...+n^2) \nonumber \\
 &\Rightarrow s_{n} < \frac{b^3}{3} < S_{n} \nonumber
\end{align}
where $s_{n}$ and $S_{n}$ are the lower and upper approximation (rectangles), respectively, for the area under the parabola. We then have to show that $A=\frac{b^3}{3}$ is the only number that satisfies 
\begin{equation}
 s_{n}<A<S_{n} \label{2}
\end{equation}
for every $n\geq1$. Using the left-most side of the first inequality, he adds $n^2$ and then multiplies both sides by $\frac{b^3}{n^3}$,
\begin{align*}
 &\frac{b^3}{n^3}(1^2+2^2+...+n^2)<\frac{b^3}{3}+\frac{b^3}{n^2} \\
 &\Rightarrow S_{n}<\frac{b^3}{3}+\frac{b^3}{n^2}
\end{align*}
for the right-most side of the inequality, he subtracts $n^2$ and multiplies by $\frac{b^3}{n^3}$,
\begin{align*}
 &\frac{b^3}{3}-\frac{b^3}{n^2}<\frac{b^3}{n^3}(1^2+2^2+...+(n-1)^2)\\
 &\Rightarrow \frac{b^3}{3}-\frac{b^3}{n^3}<s_{n} 
\end{align*}
this implies, 
\begin{align*}
 \frac{b^3}{3}-\frac{b^3}{n^2}<A<\frac{b^3}{3}+\frac{b^3}{n^2}
\end{align*}
This is where it gets confusing.  He says the only possibilities are:
$$\begin{array}{ccc}
 A>\frac{b^3}{3},& A<\frac{b^3}{3},& A=\frac{b^3}{3}
\end{array}$$
and proceeds to show that $A=\frac{b^3}{3}$ via contradictions for the first two cases. This is fine, but what about $\frac{b^3}{n^2}$ in the inequality? why don't we have to consider the possible relationships between $A$ and $\frac{b^3}{n^2}$?

First, there is a typo in your question. The inequality from the book is  
$$\frac{b^3}{3} - \frac{b^3}{n} < A < \frac{b^3}{3} + \frac{b^3}{n}$$
Picking it up from here. Note that the Apostol's ultimate goal in this proof is to somehow show that $A = \frac{b^3}{3}$. He basically just applied the law of trichotomy which says (I write informally) that given any two arbitrary  numbers $x$ and $y$, exactly one of the following is true: $x < y$ or $x > y$ or $x = y$.  
Applying that, we have $A$ which is an arbitrary number and we have $\frac{b^3}{3}$ also a number. We do not have any idea about the relationship between $A$ and $\frac{b^3}{3}$. However, we know from the law of Trichotomy that exactly one of the following (relations) is true: $A < \frac{b^3}{3}$ or $A > \frac{b^3}{3}$ or $A = \frac{b^3}{3}$.
If we're able to show that the first two cases are false, then the third case must be true which is exactly what he did.
