In `triangleABC`, If `r_(1), r_(2), r_(3)` are exradii opposites to angles A,B and C respectively. Then `(1/r_(1) +1/r_(2)) (1/r_(2)+1/r_(3)) (1/r_(3)+1/r_(1))` is equal to

Options

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

Answer

To solve the problem, we need to find the value of the expression:

\[ (1/r_1 + 1/r_2)(1/r_2 + 1/r_3)(1/r_3 + 1/r_1) \]

where \( r_1, r_2, r_3 \) are the exradii of triangle \( ABC \) opposite to angles \( A, B, \) and \( C \) respectively.

### Step 1: Express the exradii in terms of the area and semi-perimeter

The exradii \( r_1, r_2, r_3 \) can be expressed as:

\[ r_1 = \frac{\Delta}{s - a}, \quad r_2 = \frac{\Delta}{s - b}, \quad r_3 = \frac{\Delta}{s - c} \]

where \( \Delta \) is the area of the triangle and \( s \) is the semi-perimeter given by \( s = \frac{a + b + c}{2} \).

### Step 2: Find the reciprocals of the exradii

Calculating the reciprocals, we have:

\[ \frac{1}{r_1} = \frac{s - a}{\Delta}, \quad \frac{1}{r_2} = \frac{s - b}{\Delta}, \quad \frac{1}{r_3} = \frac{s - c}{\Delta} \]

### Step 3: Substitute into the expression

Now substituting these values into the expression:

\[ (1/r_1 + 1/r_2) = \left(\frac{s - a}{\Delta} + \frac{s - b}{\Delta}\right) = \frac{(s - a) + (s - b)}{\Delta} = \frac{2s - (a + b)}{\Delta} \]

Similarly,

\[ (1/r_2 + 1/r_3) = \frac{2s - (b + c)}{\Delta} \]

\[ (1/r_3 + 1/r_1) = \frac{2s - (c + a)}{\Delta} \]

### Step 4: Multiply the expressions

Now we multiply these three results together:

\[ (1/r_1 + 1/r_2)(1/r_2 + 1/r_3)(1/r_3 + 1/r_1) = \left(\frac{2s - (a + b)}{\Delta}\right) \left(\frac{2s - (b + c)}{\Delta}\right) \left(\frac{2s - (c + a)}{\Delta}\right) \]

### Step 5: Simplify the expression

This can be simplified as:

\[ = \frac{(2s - (a + b))(2s - (b + c))(2s - (c + a))}{\Delta^3} \]

### Step 6: Substitute \( 2s \)

Since \( 2s = a + b + c \), we can express:

\[ 2s - (a + b) = c, \quad 2s - (b + c) = a, \quad 2s - (c + a) = b \]

Thus, we have:

\[ = \frac{abc}{\Delta^3} \]

### Step 7: Use the area formula

We know that the area \( \Delta \) of triangle \( ABC \) can be expressed using the formula:

\[ \Delta = \frac{abc}{4R} \]

where \( R \) is the circumradius. Therefore, substituting this into our expression gives:

\[ \Delta^3 = \left(\frac{abc}{4R}\right)^3 = \frac{a^3b^3c^3}{64R^3} \]

### Step 8: Final expression

Substituting back, we get:

\[ (1/r_1 + 1/r_2)(1/r_2 + 1/r_3)(1/r_3 + 1/r_1) = \frac{abc}{\frac{a^3b^3c^3}{64R^3}} = \frac{64R^3}{abc} \]

Thus, the final result is:

\[ \boxed{64R^3} \]