Question:

Step 1: The given determinant is $$ D = \begin{vmatrix} (x^2+1)^2 & (xy+1)^2 & (xz+1)^2 \\ (xy+1)^2 & (y^2+1)^2 & (yz+1)^2 \\ (xz+1)^2 & (yz+1)^2 & (z^2+1)^2 \end{vmatrix} $$ Step 2: The elements of the determinant are of the form $(a_i b_j + c_i d_j)^2$. Let us define the following values: $a_1=x, a_2=y, a_3=z$ $c_1=1, c_2=1, c_3=1$ $b_1=x, b_2=y, b_3=z$ $d_1=1, d_2=1, d_3=1$ With these definitions, the general element $M_{ij}$ of the determinant can be expressed as $(a_i b_j + c_i d_j)^2$. For example: $M_{11} = (a_1 b_1 + c_1 d_1)^2 = (x \cdot x + 1 \cdot 1)^2 = (x^2+1)^2$ $M_{12} = (a_1 b_2 + c_1 d_2)^2 = (x \cdot y + 1 \cdot 1)^2 = (xy+1)^2$ This confirms that the given determinant matches this form. Step 3: Apply the determinant identity for this specific form. For a $3 \times 3$ matrix with entries $M_{ij} = (a_i b_j + c_i d_j)^2$, the determinant is given by the identity: $$ \det(M) = 2 (a_1c_2-a_2c_1)(a_2c_3-a_3c_2)(a_3c_1-a_1c_3) \cdot (b_1d_2-b_2d_1)(b_2d_3-b_3d_2)(b_3d_1-b_1d_3) $$ Step 4: Substitute the defined values into the identity. First, calculate the terms involving $a_i$ and $c_i$: $a_1c_2-a_2c_1 = (x \cdot 1 - y \cdot 1) = x-y$ $a_2c_3-a_3c_2 = (y \cdot 1 - z \cdot 1) = y-z$ $a_3c_1-a_1c_3 = (z \cdot 1 - x \cdot 1) = z-x$ The product of these terms is $(x-y)(y-z)(z-x)$. Next, calculate the terms involving $b_j$ and $d_j$: $b_1d_2-b_2d_1 = (x \cdot 1 - y \cdot 1) = x-y$ $b_2d_3-b_3d_2 = (y \cdot 1 - z \cdot 1) = y-z$ $b_3d_1-b_1d_3 = (z \cdot 1 - x \cdot 1) = z-x$ The product of these terms is also $(x-y)(y-z)(z-x)$. Step 5: Combine the terms and determine the value of $k$. Substitute these products back into the determinant identity: $$ D = 2 \left[ (x-y)(y-z)(z-x) \right] \left[ (x-y)(y-z)(z-x) \right] $$ $$ D = 2 (x-y)^2(y-z)^2(z-x)^2 $$ The problem states that the determinant is equal to $k(x-y)^2(y-z)^2(z-x)^2$. By comparing our result with the given form, we find that $k=2$.