Let M and m respectively be the maximum and the minimum value of
$$f(x) =\begin{vmatrix}\mathbf{1+\sin^{2}x} & \mathbf{\cos^{2}x} & \mathbf{4\sin 4x} \\\mathbf{\sin^{2}x} &\mathbf{1+\cos^{2}x} & \mathbf{4\sin 4x} \\\mathbf{\sin^{2}x} &\mathbf{\cos^{2}x} & \mathbf{1+4\sin 4x}\end{vmatrix}$$, $$x \in R$$ then $$M^{4}-m^{4}$$ is equal to :

The given function is:

$$f(x) = \begin{vmatrix} 1+\sin^{2}x & \cos^{2}x & 4\sin 4x \\ \sin^{2}x & 1+\cos^{2}x & 4\sin 4x \\ \sin^{2}x & \cos^{2}x & 1+4\sin 4x \end{vmatrix}$$

To simplify the determinant, perform row operations. Subtract row 3 from row 1 and row 2:

$$R_1 - R_3 = \begin{bmatrix} (1+\sin^{2}x) - \sin^{2}x & \cos^{2}x - \cos^{2}x & 4\sin 4x - (1+4\sin 4x) \end{bmatrix} = \begin{bmatrix} 1 & 0 & -1 \end{bmatrix}$$

$$R_2 - R_3 = \begin{bmatrix} \sin^{2}x - \sin^{2}x & (1+\cos^{2}x) - \cos^{2}x & 4\sin 4x - (1+4\sin 4x) \end{bmatrix} = \begin{bmatrix} 0 & 1 & -1 \end{bmatrix}$$

The determinant remains unchanged, so:

$$f(x) = \begin{vmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ \sin^{2}x & \cos^{2}x & 1+4\sin 4x \end{vmatrix}$$

Expand along the first row:

$$f(x) = 1 \cdot \begin{vmatrix} 1 & -1 \\ \cos^{2}x & 1+4\sin 4x \end{vmatrix} - 0 \cdot \begin{vmatrix} 0 & -1 \\ \sin^{2}x & 1+4\sin 4x \end{vmatrix} + (-1) \cdot \begin{vmatrix} 0 & 1 \\ \sin^{2}x & \cos^{2}x \end{vmatrix}$$

The second term is zero. Compute the other determinants:

First determinant: $$\begin{vmatrix} 1 & -1 \\ \cos^{2}x & 1+4\sin 4x \end{vmatrix} = 1 \cdot (1+4\sin 4x) - (-1) \cdot \cos^{2}x = 1 + 4\sin 4x + \cos^{2}x$$

Second determinant: $$\begin{vmatrix} 0 & 1 \\ \sin^{2}x & \cos^{2}x \end{vmatrix} = 0 \cdot \cos^{2}x - 1 \cdot \sin^{2}x = -\sin^{2}x$$

So:

$$f(x) = (1 + 4\sin 4x + \cos^{2}x) - (-\sin^{2}x) = 1 + 4\sin 4x + \cos^{2}x + \sin^{2}x$$

Using the identity $$\sin^{2}x + \cos^{2}x = 1$$:

$$f(x) = 1 + 4\sin 4x + 1 = 2 + 4\sin 4x$$

The range of $$\sin 4x$$ is $$[-1, 1]$$, so:

Maximum value of $$f(x)$$ occurs when $$\sin 4x = 1$$:

$$M = 2 + 4(1) = 6$$

Minimum value of $$f(x)$$ occurs when $$\sin 4x = -1$$:

$$m = 2 + 4(-1) = -2$$

Now compute $$M^4 - m^4$$:

$$M^4 = 6^4 = 1296$$

$$m^4 = (-2)^4 = 16$$

$$M^4 - m^4 = 1296 - 16 = 1280$$

Alternatively, using the difference of squares:

$$M^4 - m^4 = (M^2)^2 - (m^2)^2 = (M^2 - m^2)(M^2 + m^2)$$

$$M^2 = 36, \quad m^2 = 4$$

$$M^2 - m^2 = 36 - 4 = 32$$

$$M^2 + m^2 = 36 + 4 = 40$$

$$M^4 - m^4 = 32 \times 40 = 1280$$

The value of $$M^4 - m^4$$ is 1280.