If `f(x) = |(1,x,x+1),(2x,x(x-1),x(x+1)),(3x(x-1),x(x-1)(x-2),x(x+1)(x-1))|`, using properties of determinant, find f(2x) - f(x).

Answer

To solve the problem, we need to evaluate the determinant function \( f(x) \) and find \( f(2x) - f(x) \).

### Step 1: Write down the determinant function
We have:
\[ f(x) = \begin{vmatrix} 1 & 2x & 3x \\ x & x(x-1) & x(x+1) \\ x(x-1) & x(x-1)(x-2) & x(x+1)(x-1) \end{vmatrix} \]

### Step 2: Factor out common terms from the columns
From the second column, we can factor out \( x \):
\[ f(x) = \begin{vmatrix} 1 & 2x & 3x \\ x & x(x-1) & x(x+1) \\ x(x-1) & x(x-1)(x-2) & x(x+1)(x-1) \end{vmatrix} = x \cdot \begin{vmatrix} 1 & 2 & 3 \\ 1 & (x-1) & (x+1) \\ (x-1) & (x-1)(x-2) & (x+1)(x-1) \end{vmatrix} \]

### Step 3: Factor out common terms from the third column
Now, we can factor out \( (x+1) \) from the third column:
\[ f(x) = x(x+1) \cdot \begin{vmatrix} 1 & 2 & 3 \\ 1 & (x-1) & 1 \\ (x-1) & (x-1)(x-2) & (x-1) \end{vmatrix} \]

### Step 4: Simplify the determinant
Next, we can simplify the determinant by performing row operations. We can subtract the first row from the second and third rows:
\[ f(x) = x(x+1) \cdot \begin{vmatrix} 1 & 2 & 3 \\ 0 & (x-1)-2 & 1-3 \\ 0 & (x-1)(x-2)-(x-1) & (x-1)-3 \end{vmatrix} \]
This simplifies to:
\[ f(x) = x(x+1) \cdot \begin{vmatrix} 1 & 2 & 3 \\ 0 & x-3 & -2 \\ 0 & (x-1)(x-2)-(x-1) & -2 \end{vmatrix} \]

### Step 5: Evaluate the determinant
The determinant of a matrix with a row of zeros is zero:
\[ f(x) = 0 \]

### Step 6: Evaluate \( f(2x) \)
Since \( f(x) = 0 \), it follows that:
\[ f(2x) = 0 \]

### Step 7: Calculate \( f(2x) - f(x) \)
Now, we can find:
\[ f(2x) - f(x) = 0 - 0 = 0 \]

### Final Answer
Thus, the final result is:
\[ f(2x) - f(x) = 0 \]