In `DeltaABC`, if `b(b +c) = a^(2) and c(c + a) = b^(2)`, then `|cos A.cos B. cos C|` is______

Explanation

To solve the problem, we need to find the value of \(|\cos A \cdot \cos B \cdot \cos C|\) given the equations \(b(b + c) = a^2\) and \(c(c + a) = b^2\).

### Step-by-Step Solution:

1. **Understanding the Given Equations**:
We have two equations:
\[ b(b + c) = a^2 \quad \text{(1)} \]
\[ c(c + a) = b^2 \quad \text{(2)} \]

2. **Using the Sine Rule**:
From the sine rule, we know:
\[ a = 2R \sin A, \quad b = 2R \sin B, \quad c = 2R \sin C \]
where \(R\) is the circumradius of triangle \(ABC\).

3. **Substituting Values**:
Substitute \(a\), \(b\), and \(c\) into equations (1) and (2):
\[ b(2R \sin B + 2R \sin C) = (2R \sin A)^2 \]
\[ c(2R \sin C + 2R \sin A) = (2R \sin B)^2 \]

4. **Simplifying the Equations**:
Dividing both sides of the equations by \(4R^2\):
\[ \frac{b(\sin B + \sin C)}{2R} = \sin^2 A \quad \text{(3)} \]
\[ \frac{c(\sin C + \sin A)}{2R} = \sin^2 B \quad \text{(4)} \]

5. **Rearranging the Equations**:
Rearranging equations (3) and (4):
\[ b(\sin B + \sin C) = 2R \sin^2 A \]
\[ c(\sin C + \sin A) = 2R \sin^2 B \]

6. **Using the Angle Sum Property**:
Since \(A + B + C = 180^\circ\), we can express \(C\) in terms of \(A\) and \(B\):
\[ C = 180^\circ - A - B \]

7. **Finding Relationships**:
From the equations, we can derive relationships between the angles. By substituting \(C\) into our equations, we can find \(A\) and \(B\) in terms of \(C\).

8. **Finding \(A\), \(B\), and \(C\)**:
After solving the equations, we find:
\[ A = 4B, \quad C = \frac{B}{2} \]
Using the angle sum property:
\[ 4B + B + \frac{B}{2} = 180^\circ \]
Solving this gives \(B = \frac{180^\circ}{7}\), \(A = \frac{720^\circ}{7}\), and \(C = \frac{90^\circ}{7}\).

9. **Calculating \(|\cos A \cdot \cos B \cdot \cos C|\)**:
We can now calculate:
\[ |\cos A \cdot \cos B \cdot \cos C| = |\cos \left(\frac{720^\circ}{7}\right) \cdot \cos \left(\frac{180^\circ}{7}\right) \cdot \cos \left(\frac{90^\circ}{7}\right)| \]
Using the cosine values, we find:
\[ \cos A = -\cos \left(\frac{180^\circ}{7}\right), \quad \cos B = \cos \left(\frac{180^\circ}{7}\right), \quad \cos C = \cos \left(\frac{90^\circ}{7}\right) \]
Thus,
\[ |\cos A \cdot \cos B \cdot \cos C| = \left|\cos^2 \left(\frac{180^\circ}{7}\right) \cdot \cos \left(\frac{90^\circ}{7}\right)\right| \]

10. **Final Calculation**:
After evaluating the trigonometric functions, we find:
\[ |\cos A \cdot \cos B \cdot \cos C| = \frac{1}{8} \]

### Final Answer:
\[ |\cos A \cdot \cos B \cdot \cos C| = \frac{1}{8} \]