In a triangle `{:abs((a,b,c),(b,c,a),(c,a,b)):}=0 " then " sin^3 A+sin^3 B+sin^3C=.....`

Answer

To solve the problem, we need to analyze the given determinant condition and apply the sine rule in the context of a triangle. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Understanding the Determinant Condition**:
We start with the determinant condition given by:
\[ \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0 \]
This means that the rows (or columns) of the determinant are linearly dependent.

**Hint**: Recall that if a determinant is zero, it indicates a linear relationship between the rows or columns.

2. **Column Operation**:
We perform the column operation: \( C_1 \to C_1 + C_2 + C_3 \). This gives us:
\[ \begin{vmatrix} a+b+c & b & c \\ a+b+c & c & a \\ a+b+c & a & b \end{vmatrix} = 0 \]
Now, we can factor out \( (a+b+c) \) from the first column:
\[ (a+b+c) \begin{vmatrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{vmatrix} = 0 \]

**Hint**: Remember that factoring out a common term from a column in a determinant does not change its value.

3. **Setting Up the Equation**:
Since the determinant is zero, we have two cases:
- \( a + b + c = 0 \)
- The determinant \( \begin{vmatrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{vmatrix} = 0 \)

**Hint**: Think about the implications of \( a + b + c = 0 \) in terms of the angles of the triangle.

4. **Applying the Sine Rule**:
According to the sine rule, we have:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k \]
From this, we can express \( a, b, c \) in terms of \( k \) and the sine of the angles:
\[ a = k \sin A, \quad b = k \sin B, \quad c = k \sin C \]

**Hint**: The sine rule relates the sides of a triangle to the sines of its angles.

5. **Finding \( \sin^3 A + \sin^3 B + \sin^3 C \)**:
We substitute \( a, b, c \) into the expression \( \sin^3 A + \sin^3 B + \sin^3 C \):
\[ \sin^3 A + \sin^3 B + \sin^3 C = \frac{a^3}{k^3} + \frac{b^3}{k^3} + \frac{c^3}{k^3} \]
This can be simplified as:
\[ = \frac{1}{k^3} (a^3 + b^3 + c^3) \]

**Hint**: Remember the identity for the sum of cubes: \( x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - xz - yz) \).

6. **Using the Condition \( a + b + c = 0 \)**:
Since \( a + b + c = 0 \), we can use the identity:
\[ a^3 + b^3 + c^3 = 3abc \]
Therefore:
\[ \sin^3 A + \sin^3 B + \sin^3 C = \frac{3abc}{k^3} \]

**Hint**: The relationship between the sides and angles in a triangle is crucial for simplifying expressions.

### Final Result:
Thus, we conclude that:
\[ \sin^3 A + \sin^3 B + \sin^3 C = 3 \sin A \sin B \sin C \]