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}<A<\frac{b^3}{3}+\frac{b^3}{n} \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 $\frac{b^3}{n}$ in the inequality? why don't we have to consider the possible relationships between $A$ and $\frac{b^3}{n}$?
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.