`a(cos B cos C + cosA)=b(cos C cosA+cosB)=c(cos A cos B +cos C)`

Answer

To prove the equality \( a(\cos B \cos C + \cos A) = b(\cos C \cos A + \cos B) = c(\cos A \cos B + \cos C) \), we will follow a systematic approach.

### Step-by-Step Solution

1. **Start with the given equations:**
\[ a(\cos B \cos C + \cos A) = b(\cos C \cos A + \cos B) = c(\cos A \cos B + \cos C) \]

2. **Set the first two expressions equal:**
\[ a(\cos B \cos C + \cos A) = b(\cos C \cos A + \cos B) \]

3. **Rearranging the equation:**
\[ a \cos B \cos C + a \cos A = b \cos C \cos A + b \cos B \]

4. **Rearranging terms:**
\[ a \cos B \cos C - b \cos B + a \cos A - b \cos C \cos A = 0 \]

5. **Factor out common terms:**
\[ \cos B (a \cos C - b) + \cos A (a - b \cos C) = 0 \]

6. **Now, set the second and third expressions equal:**
\[ b(\cos C \cos A + \cos B) = c(\cos A \cos B + \cos C) \]

7. **Rearranging the equation:**
\[ b \cos C \cos A + b \cos B = c \cos A \cos B + c \cos C \]

8. **Rearranging terms:**
\[ b \cos C \cos A - c \cos C + b \cos B - c \cos A \cos B = 0 \]

9. **Factor out common terms:**
\[ \cos C (b \cos A - c) + \cos B (b - c \cos A) = 0 \]

10. **Now, set the first and third expressions equal:**
\[ a(\cos B \cos C + \cos A) = c(\cos A \cos B + \cos C) \]

11. **Rearranging the equation:**
\[ a \cos B \cos C + a \cos A = c \cos A \cos B + c \cos C \]

12. **Rearranging terms:**
\[ a \cos B \cos C - c \cos C + a \cos A - c \cos A \cos B = 0 \]

13. **Factor out common terms:**
\[ \cos C (a \cos B - c) + \cos A (a - c \cos B) = 0 \]

14. **Conclusion:**
Since all three equations can be expressed in the form \( P + Q = 0 \) where \( P \) and \( Q \) are functions of the angles and sides of the triangle, we conclude that:
\[ a(\cos B \cos C + \cos A) = b(\cos C \cos A + \cos B) = c(\cos A \cos B + \cos C) \]