Suppose the side lengths of triangle $ABC$ are the roots of polynomial $x^3 - 27x^2 + 222x - 540$. What is the product of its inradius and circumradius?
1. **Identify the roots of the polynomial:**
   The polynomial given is \(x^3 - 27x^2 + 222x - 540\). The roots of this polynomial are the side lengths \(a\), \(b\), and \(c\) of the triangle \(ABC\).

2. **Sum and product of the roots:**
   By Vieta's formulas, for the polynomial \(x^3 - 27x^2 + 222x - 540\):
   - The sum of the roots \(a + b + c = 27\).
   - The product of the roots \(abc = 540\).

3. **Inradius and circumradius formulas:**
   - The inradius \(r\) of a triangle is given by:
     \[
     r = \frac{\text{Area}}{s}
     \]
     where \(s\) is the semi-perimeter of the triangle, \(s = \frac{a + b + c}{2}\).
   - The circumradius \(R\) of a triangle is given by:
     \[
     R = \frac{abc}{4 \cdot \text{Area}}
     \]

4. **Product of inradius and circumradius:**
   - The product of the inradius and circumradius is:
     \[
     r \cdot R = \left(\frac{\text{Area}}{s}\right) \cdot \left(\frac{abc}{4 \cdot \text{Area}}\right)
     \]
     Simplifying this expression:
     \[
     r \cdot R = \frac{\text{Area} \cdot abc}{s \cdot 4 \cdot \text{Area}} = \frac{abc}{4s}
     \]

5. **Substitute the known values:**
   - The semi-perimeter \(s\) is:
     \[
     s = \frac{a + b + c}{2} = \frac{27}{2}
     \]
   - The product of the roots \(abc = 540\).

6. **Calculate the product \(r \cdot R\):**
   \[
   r \cdot R = \frac{abc}{4s} = \frac{540}{4 \cdot \frac{27}{2}} = \frac{540}{54} = 10
   \]

The final answer is \(\boxed{10}\).