Let `f(x)=|(a^2+(b^2+c^2)cosx,ab(1-cosx),ac(1-cosx)),(ba(1-cosx),b^2+(c^2+a^2)cosx,bc(1-cosx)),(ca(1-cosx),cb(1-cosx),c^2+(a^2+b^2)cosx)|` where `"x"in[-pi/2,pi/2]` and a,b,c
`f(x) = `

Options

A. `(suma^2)cosx`
B. `(suma^2)^2cos^2x`
C. `(suma^2)^3cos^2x`
D. None

Answer

To solve the problem, we need to evaluate the determinant given in the function \( f(x) \) and find its maximum value over the interval \( x \in [-\frac{\pi}{2}, \frac{\pi}{2}] \).

### Step-by-Step Solution:

1. **Write the Determinant**:
The function is given as:
\[ f(x) = \begin{vmatrix} a^2 + (b^2 + c^2) \cos x & ab(1 - \cos x) & ac(1 - \cos x) \\ ba(1 - \cos x) & b^2 + (c^2 + a^2) \cos x & bc(1 - \cos x) \\ ca(1 - \cos x) & cb(1 - \cos x) & c^2 + (a^2 + b^2) \cos x \end{vmatrix} \]

2. **Factor Out Constants**:
We can factor out \( abc \) from the determinant:
\[ f(x) = \frac{1}{abc} \begin{vmatrix} a^3 + a(b^2 + c^2) \cos x & ab^2(1 - \cos x) & ac^2(1 - \cos x) \\ b^2a(1 - \cos x) & b^3 + b(c^2 + a^2) \cos x & bc^2(1 - \cos x) \\ c^2a(1 - \cos x) & cb^2(1 - \cos x) & c^3 + c(a^2 + b^2) \cos x \end{vmatrix} \]

3. **Simplify the Determinant**:
We can perform row operations to simplify the determinant. Subtract the first row from the second and third rows:
\[ R_2 \rightarrow R_2 - R_1, \quad R_3 \rightarrow R_3 - R_1 \]
This gives us:
\[ f(x) = \frac{1}{abc} \begin{vmatrix} a^2 + (b^2 + c^2) \cos x & ab(1 - \cos x) & ac(1 - \cos x) \\ 0 & b^2 + (c^2 + a^2) \cos x - (a^2 + (b^2 + c^2) \cos x) & bc(1 - \cos x) - ab(1 - \cos x) \\ 0 & cb(1 - \cos x) - ac(1 - \cos x) & c^2 + (a^2 + b^2) \cos x - (a^2 + (b^2 + c^2) \cos x) \end{vmatrix} \]

4. **Evaluate the Determinant**:
After simplification, we find:
\[ f(x) = (a^2 + b^2 + c^2)^2 \cos x \]

5. **Find the Maximum Value**:
The maximum value of \( f(x) \) occurs when \( \cos x \) is maximized. Since \( \cos x \) reaches its maximum value of 1 at \( x = 0 \):
\[ f(0) = (a^2 + b^2 + c^2)^2 \]

### Conclusion:
The maximum value of \( f(x) \) is:
\[ f(x) = (a^2 + b^2 + c^2)^2 \]