If `cosA+cosB+cosC=0, then cos3A+cos3B+cos3C` is equal to

Options

A. `cosA cosBcosC`
B. `12cosAcosBcosC`
C. 0
D. `8cos^(2)Acos^(3)Bcos^(3)C`

Answer

To solve the problem, we need to find the value of \( \cos 3A + \cos 3B + \cos 3C \) given that \( \cos A + \cos B + \cos C = 0 \).

### Step-by-step Solution:

1. **Use the formula for \( \cos 3\theta \)**:
The formula for \( \cos 3\theta \) is:
\[ \cos 3\theta = 4 \cos^3 \theta - 3 \cos \theta \]
Therefore, we can express \( \cos 3A \), \( \cos 3B \), and \( \cos 3C \) as:
\[ \cos 3A = 4 \cos^3 A - 3 \cos A \]
\[ \cos 3B = 4 \cos^3 B - 3 \cos B \]
\[ \cos 3C = 4 \cos^3 C - 3 \cos C \]

2. **Combine the expressions**:
Now, we can combine these:
\[ \cos 3A + \cos 3B + \cos 3C = (4 \cos^3 A + 4 \cos^3 B + 4 \cos^3 C) - (3 \cos A + 3 \cos B + 3 \cos C) \]
This simplifies to:
\[ = 4 (\cos^3 A + \cos^3 B + \cos^3 C) - 3 (\cos A + \cos B + \cos C) \]

3. **Substitute the known value**:
From the problem, we know that \( \cos A + \cos B + \cos C = 0 \). Therefore, the second term becomes:
\[ -3 (\cos A + \cos B + \cos C) = -3 \cdot 0 = 0 \]
Thus, we have:
\[ \cos 3A + \cos 3B + \cos 3C = 4 (\cos^3 A + \cos^3 B + \cos^3 C) \]

4. **Use the identity for the sum of cubes**:
We can use the identity:
\[ \cos^3 A + \cos^3 B + \cos^3 C = 3 \cos A \cos B \cos C \]
when \( \cos A + \cos B + \cos C = 0 \). Therefore, we can substitute this into our expression:
\[ \cos 3A + \cos 3B + \cos 3C = 4 \cdot 3 \cos A \cos B \cos C = 12 \cos A \cos B \cos C \]

5. **Final Result**:
Thus, the final result is:
\[ \cos 3A + \cos 3B + \cos 3C = 12 \cos A \cos B \cos C \]

### Conclusion:
The value of \( \cos 3A + \cos 3B + \cos 3C \) is \( 12 \cos A \cos B \cos C \).