Let `f(x)=|(1,3,5),(x-2,3x^(2)-12,5x^(3)-40),(x-3,3x^(2)-27,2x^(3)-54)|` then `f(2)f(3)+f(2)f(7)+f(3)f(7)=`__________

Answer

To solve the problem, we need to evaluate the function \( f(x) \) which is defined as the determinant of the matrix:

\[ f(x) = \begin{vmatrix} 1 & 3 & 5 \\ x-2 & 3x^2 - 12 & 5x^3 - 40 \\ x-3 & 3x^2 - 27 & 2x^3 - 54 \end{vmatrix} \]

We will calculate \( f(2) \), \( f(3) \), and \( f(7) \) and then find the expression \( f(2)f(3) + f(2)f(7) + f(3)f(7) \).

### Step 1: Calculate \( f(2) \)

Substituting \( x = 2 \) into the matrix:

\[ f(2) = \begin{vmatrix} 1 & 3 & 5 \\ 2-2 & 3(2^2) - 12 & 5(2^3) - 40 \\ 2-3 & 3(2^2) - 27 & 2(2^3) - 54 \end{vmatrix} \]

This simplifies to:

\[ f(2) = \begin{vmatrix} 1 & 3 & 5 \\ 0 & 12 - 12 & 40 - 40 \\ -1 & 12 - 27 & 16 - 54 \end{vmatrix} \]

Calculating the second row:

\[ f(2) = \begin{vmatrix} 1 & 3 & 5 \\ 0 & 0 & 0 \\ -1 & -15 & -38 \end{vmatrix} \]

Since one row is entirely zeros, the determinant is:

\[ f(2) = 0 \]

### Step 2: Calculate \( f(3) \)

Now substituting \( x = 3 \):

\[ f(3) = \begin{vmatrix} 1 & 3 & 5 \\ 3-2 & 3(3^2) - 12 & 5(3^3) - 40 \\ 3-3 & 3(3^2) - 27 & 2(3^3) - 54 \end{vmatrix} \]

This simplifies to:

\[ f(3) = \begin{vmatrix} 1 & 3 & 5 \\ 1 & 27 - 12 & 135 - 40 \\ 0 & 27 - 27 & 54 - 54 \end{vmatrix} \]

Calculating the second row:

\[ f(3) = \begin{vmatrix} 1 & 3 & 5 \\ 1 & 15 & 95 \\ 0 & 0 & 0 \end{vmatrix} \]

Again, since one row is entirely zeros, the determinant is:

\[ f(3) = 0 \]

### Step 3: Calculate \( f(7) \)

Now substituting \( x = 7 \):

\[ f(7) = \begin{vmatrix} 1 & 3 & 5 \\ 7-2 & 3(7^2) - 12 & 5(7^3) - 40 \\ 7-3 & 3(7^2) - 27 & 2(7^3) - 54 \end{vmatrix} \]

This simplifies to:

\[ f(7) = \begin{vmatrix} 1 & 3 & 5 \\ 5 & 147 - 12 & 1715 - 40 \\ 4 & 147 - 27 & 686 - 54 \end{vmatrix} \]

Calculating the second row:

\[ f(7) = \begin{vmatrix} 1 & 3 & 5 \\ 5 & 135 & 1675 \\ 4 & 120 & 632 \end{vmatrix} \]

We will compute this determinant, but since \( f(2) \) and \( f(3) \) are both zero, we can directly evaluate:

### Step 4: Calculate the final expression

Now we compute:

\[ f(2)f(3) + f(2)f(7) + f(3)f(7) = 0 \cdot 0 + 0 \cdot f(7) + 0 \cdot f(7) = 0 \]

Thus, the final answer is:

\[ \boxed{0} \]