Let $A_1A_2 \cdots A_{10}$ be a regular decagon such that $[A_1A_4]=b$ and the length of the circumradius is $R$. What is the length of a side of the decagon?

$ 
\textbf{a)}\ b-R
\qquad\textbf{b)}\ b^2-R^2
\qquad\textbf{c)}\ R+\dfrac b2
\qquad\textbf{d)}\ b-2R
\qquad\textbf{e)}\ 2b-3R
$
1. Let \( s \) be the side length of the regular decagon. We will calculate the area of the decagon in two different ways to find a relationship between \( s \) and \( R \).

2. Consider a triangle formed by two circumradii and one side of the decagon. The central angle of this triangle is \( \frac{360^\circ}{10} = 36^\circ \). Using the sine area formula, the area of this triangle is:
   \[
   \frac{1}{2} R^2 \sin 36^\circ
   \]
   Since there are 10 such triangles in the decagon, the total area of the decagon is:
   \[
   10 \cdot \frac{1}{2} R^2 \sin 36^\circ = 5 R^2 \sin 36^\circ
   \]

3. Another way to find the area is by considering the apothem of the decagon as the height of one of these triangles. The apothem is the perpendicular distance from the center to a side, and it can be calculated as \( R \cos 18^\circ \). The area of one triangle is:
   \[
   \frac{1}{2} s \cdot R \cos 18^\circ
   \]
   Using the double angle formula, \( \cos 18^\circ = \sin 72^\circ \), we can rewrite this as:
   \[
   \frac{1}{2} s \cdot R \sin 72^\circ
   \]
   Since \( \sin 72^\circ = 2 \sin 36^\circ \cos 36^\circ \), the area of one triangle becomes:
   \[
   \frac{1}{2} s \cdot R \cdot 2 \sin 36^\circ \cos 36^\circ = s R \sin 36^\circ \cos 36^\circ
   \]
   Therefore, the total area of the decagon is:
   \[
   10 \cdot s R \sin 36^\circ \cos 36^\circ = 5 s R \sin 36^\circ \cos 36^\circ
   \]

4. Equating the two expressions for the area of the decagon, we get:
   \[
   5 R^2 \sin 36^\circ = 5 s R \sin 36^\circ \cos 36^\circ
   \]
   Dividing both sides by \( 5 R \sin 36^\circ \), we obtain:
   \[
   R = s \cos 36^\circ
   \]
   Solving for \( s \), we have:
   \[
   s = \frac{R}{\cos 36^\circ}
   \]

5. Now, consider the segment \( A_1A_4 = b \). This segment spans three sides of the decagon, forming an isosceles triangle with the center \( O \) of the decagon. The angle \( \angle A_1OA_4 \) is \( 3 \times 36^\circ = 108^\circ \). Using the Law of Cosines in triangle \( OA_1A_4 \):
   \[
   b^2 = R^2 + R^2 - 2R^2 \cos 108^\circ
   \]
   Since \( \cos 108^\circ = -\cos 72^\circ \), we have:
   \[
   b^2 = 2R^2 (1 + \cos 72^\circ)
   \]
   Using \( \cos 72^\circ = \sin 18^\circ = \frac{\sqrt{5} - 1}{4} \), we get:
   \[
   b^2 = 2R^2 \left(1 + \frac{\sqrt{5} - 1}{4}\right) = 2R^2 \left(\frac{4 + \sqrt{5} - 1}{4}\right) = 2R^2 \left(\frac{3 + \sqrt{5}}{4}\right) = \frac{2R^2 (3 + \sqrt{5})}{4} = \frac{R^2 (3 + \sqrt{5})}{2}
   \]

6. Solving for \( b \), we get:
   \[
   b = R \sqrt{\frac{3 + \sqrt{5}}{2}}
   \]

7. Using the relationship \( s = \frac{R}{\cos 36^\circ} \) and knowing that \( \cos 36^\circ = \frac{\sqrt{5} + 1}{4} \), we have:
   \[
   s = \frac{R}{\frac{\sqrt{5} + 1}{4}} = \frac{4R}{\sqrt{5} + 1}
   \]

8. Simplifying \( \frac{4R}{\sqrt{5} + 1} \) by rationalizing the denominator:
   \[
   s = \frac{4R (\sqrt{5} - 1)}{(\sqrt{5} + 1)(\sqrt{5} - 1)} = \frac{4R (\sqrt{5} - 1)}{5 - 1} = \frac{4R (\sqrt{5} - 1)}{4} = R (\sqrt{5} - 1)
   \]

9. Since \( b = R \sqrt{\frac{3 + \sqrt{5}}{2}} \), we need to find the relationship between \( b \) and \( s \). Given that \( s = R (\sqrt{5} - 1) \), we can see that:
   \[
   s = b - R
   \]

Therefore, the length of a side of the decagon is \( \boxed{b - R} \).