Writing an Expression for the Volume of a Spherical Shell The question is as follows:

Write an expression for the volume of the spherical shell formed between two concentric spheres, the inner one of radius $r$, the outer one of radius $R$. Factor your answer so that it has the form $4 π · (trinomial) · (binomial)$. In this situation, what is the meaning of the binomial? What can be said about the value of the trinomial when the binomial has a very small value? Make a conjecture concerning the surface area of a sphere of radius $R$.

I know that the volume of the sphere with the inner radius of r would be $\frac{4}{3}\pi r^3$ and the volume of the sphere with the outer radius of R would be $\frac{4}{3}\pi R^3$. Therefore, the volume of the spherical shell formed between the concentric circles would be the difference between the outer sphere and the inner sphere--which, when mathematically written, would be $\frac{4}{3}\pi R^3 - \frac{4}{3}\pi r^3$. That expression, after it's factored, would be $\frac{4}{3}\pi(R^3 - r^3)$. My question is how can I represent $(R^3 - r^3)$ the product of a trinomial and binomial. Any help will be greatly appreciated! 
EDIT: I think that I may have found the answer to the question that I have presented above: $(R^2 + Rr + r^2)(R - r)$. Therefore, my complete answer to the volume of the spherical shell is $\frac{4}{3}(R^2 + Rr + r^2)(R - r)$. 
I still don't know what the binomial is representing for this particular problem, nor its relation to the trinomial when the binomial value gets smaller. I think that I can use the help for understanding what the binomial and trinomial represents as a way to formulate a conjecture regarding the surface area of a sphere with radius R. 
EDIT #2: The binomial, as I understand it, represents the thickness of the spherical shell formed by the concentric circles. 
 A: Note that $$\frac{4}{3} \pi (R^2 + Rr + r^2)(R - r)$$
has three parts. 
You have already figured out the $(R-r)$ part as the thickness of the spherical shell. 
The trinomial part is very close to $ 3r^2$ in case that the thickness is very small. 
Now when you multiply this $3r^2$ by $ \frac{4}{3} \pi$, you come up with $$ 4\pi r^2$$ which is the surface area of the shell( Approximate of course)
Thus The volume of the shell is approximately, the surface area of the shell multiplied by the thickness of the shell.  
A: Let $V(r)$ be the volume of the sphere with radius $r$ and $A(r)$ its surface area.We have
$$ V(r) = V(1)\cdot r^3$$
since concentric spheres are homothetic. In particular
$$ V(R)-V(r) = V(1)(R-r)(R^2+Rr+r^2). $$
On the other hand $A(r) = A(1)\cdot r^2$ and
$$ A(R) = \lim_{r\to R^-}\frac{V(R)-V(r)}{R-r} = \lim_{r\to R^-} V(1)(R^2+Rr+r^2)=3V(1)\,R^2.$$ 
This proves $A(1)=3\,V(1)$ and the same argument in dimension $n$ proves $A(1)=n\,V(1)$: to find the volume or the surface area of a Euclidean ball are equivalent problems. Remark: for $n=3$, the same can be shown by approximating a sphere with the union of a large number of cones and recalling that the volume of a cone is given by one third of the product between the height and the base area.
A: 
Write an expression for the volume of the spherical shell formed between two concentric spheres, the inner one of radius $r$, the outer one of radius $R$. Factor your answer so that it has the form $4 \pi \cdot (\text{trinomial}) \cdot (\text{binomial})$

$$
V(R, r) 
= \frac{4}{3}\pi (R^3-r^3) 
= \frac{4}{3}\pi(R-r)(R^2+r^2 + Rr)
= 4 \pi \, t(R,r) \, b(R,r) 
$$
with
\begin{align}
t(R,r) &= \frac{1}{3}(R^2+r^2 + Rr) \\
b(R, r) &= R-r
\end{align}

What can be said about the value of the trinomial when the binomial
  has a very small value?

In this case we have $r \approx R$ and we can estimate the volume by 
$$
V(R,r) \approx A(r) \, b(R,r)
$$
where $A(r)$ is the area of a sphere with radius $r$. So we get 
$$
t(R,r) = \frac{1}{3}(R^2+r^2 + Rr) = \frac{V(R,r)}{4\pi b(R,r)}
\approx \frac{A(r) \, b(R,r)}{4\pi b(R,r)}
= \frac{A(r)}{4\pi}
$$

Make a conjecture concerning the surface area of a sphere of radius R.

$$
A(t) \approx 4\pi\, t(R,r)
= 4\pi \, \frac{1}{3}(R^2 + r^2 + Rr) \approx 4\pi \, \frac{1}{3}(r^2 + r^2 + r^2) = 4 \pi r^2
$$
