If R be the circum radius of the triangle ABC then the value of
`R^3/((a-b)(b-c)(c-a))|(1,1,1),(sin A,sinB,sinC),(sin^2A,sin^2B,sin^2C)|` is

Options

A. 8
B. `1//8`
C. 4
D. `1//4`

Answer

To solve the problem, we need to evaluate the expression

\[ \frac{R^3}{(a-b)(b-c)(c-a)} \left| \begin{array}{ccc} 1 & 1 & 1 \\ \sin A & \sin B & \sin C \\ \sin^2 A & \sin^2 B & \sin^2 C \end{array} \right| \]

where \( R \) is the circumradius of triangle \( ABC \) and \( a, b, c \) are the lengths of the sides opposite to angles \( A, B, C \) respectively.

### Step 1: Use the relationship between sides and angles
We know that for a triangle, the circumradius \( R \) is related to the sides and angles by the formula:

\[ 2R = \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]

From this, we can express \( \sin A, \sin B, \sin C \) in terms of \( R \):

\[ \sin A = \frac{a}{2R}, \quad \sin B = \frac{b}{2R}, \quad \sin C = \frac{c}{2R} \]

### Step 2: Substitute into the determinant
Now, we substitute these values into the determinant:

\[ \left| \begin{array}{ccc} 1 & 1 & 1 \\ \frac{a}{2R} & \frac{b}{2R} & \frac{c}{2R} \\ \left(\frac{a}{2R}\right)^2 & \left(\frac{b}{2R}\right)^2 & \left(\frac{c}{2R}\right)^2 \end{array} \right| \]

This can be simplified as:

\[ \frac{1}{(2R)^2} \left| \begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{array} \right| \]

### Step 3: Factor out constants
The determinant can be factored out:

\[ \frac{1}{4R^2} \left| \begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{array} \right| \]

### Step 4: Evaluate the determinant
The determinant

\[ \left| \begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{array} \right| \]

is known to be equal to

\[ (a-b)(b-c)(c-a) \]

### Step 5: Substitute back into the expression
Now substituting this back into our expression:

\[ \frac{R^3}{(a-b)(b-c)(c-a)} \cdot \frac{1}{4R^2} \cdot (a-b)(b-c)(c-a) \]

### Step 6: Simplify
The \( (a-b)(b-c)(c-a) \) terms cancel out:

\[ \frac{R^3}{4R^2} = \frac{R}{4} \]

### Final Result
Thus, the final value is:

\[ \frac{R}{4} \]