`(1/r_1 + 1/r_2)(1/r_2 + 1/r_3)(1/r_3 + 1/r_1)=`

Options

A. `(16R)/(abc)`
B. `(R^3)/(4abc)`
C. `(64R^3)/(a^2b^2c^2)`
D. `(R^3)/(abc)`

Answer

To solve the expression \((\frac{1}{r_1} + \frac{1}{r_2})(\frac{1}{r_2} + \frac{1}{r_3})(\frac{1}{r_3} + \frac{1}{r_1})\), we will use the formulas for the radii of the incircle and circumcircle of a triangle.

### Step-by-Step Solution:

1. **Identify the Radii**:
The radii \(r_1\), \(r_2\), and \(r_3\) represent the radii of the incircle of the triangle opposite to vertices \(A\), \(B\), and \(C\) respectively. The formulas for the radii are:
\[ r_1 = \frac{\Delta}{s - a}, \quad r_2 = \frac{\Delta}{s - b}, \quad r_3 = \frac{\Delta}{s - c} \]
where \(\Delta\) is the area of the triangle and \(s\) is the semi-perimeter given by \(s = \frac{a + b + c}{2}\).

2. **Calculate \(\frac{1}{r_i}\)**:
We can express \(\frac{1}{r_i}\) for each radius:
\[ \frac{1}{r_1} = \frac{s - a}{\Delta}, \quad \frac{1}{r_2} = \frac{s - b}{\Delta}, \quad \frac{1}{r_3} = \frac{s - c}{\Delta} \]

3. **Substitute into the Expression**:
Substitute these values into the expression:
\[ \left( \frac{s - a}{\Delta} + \frac{s - b}{\Delta} \right)\left( \frac{s - b}{\Delta} + \frac{s - c}{\Delta} \right)\left( \frac{s - c}{\Delta} + \frac{s - a}{\Delta} \right) \]

4. **Simplify Each Term**:
Factor out \(\frac{1}{\Delta}\) from each term:
\[ = \frac{1}{\Delta^3} \left( (s - a + s - b)(s - b + s - c)(s - c + s - a) \right) \]
This simplifies to:
\[ = \frac{1}{\Delta^3} \left( (2s - a - b)(2s - b - c)(2s - c - a) \right) \]

5. **Substitute \(s\)**:
Recall that \(s = \frac{a + b + c}{2}\). Therefore:
\[ 2s - a - b = c, \quad 2s - b - c = a, \quad 2s - c - a = b \]
Thus, we can rewrite the expression as:
\[ = \frac{1}{\Delta^3} \cdot (c)(a)(b) = \frac{abc}{\Delta^3} \]

6. **Relate to the Circumradius**:
The area \(\Delta\) can be expressed in terms of the circumradius \(R\) as:
\[ \Delta = \frac{abc}{4R} \]
Therefore, substituting for \(\Delta\):
\[ \Delta^3 = \left(\frac{abc}{4R}\right)^3 = \frac{a^3b^3c^3}{64R^3} \]

7. **Final Expression**:
Substitute this back into the expression:
\[ = \frac{abc}{\frac{a^3b^3c^3}{64R^3}} = \frac{64R^3}{a^2b^2c^2} \]

### Final Result:
Thus, the value of the expression \((\frac{1}{r_1} + \frac{1}{r_2})(\frac{1}{r_2} + \frac{1}{r_3})(\frac{1}{r_3} + \frac{1}{r_1})\) is:
\[ \frac{64R^3}{a^2b^2c^2} \]