If `|{:((x+1),(x+1)^2,(x+1)^3),((x+2),(x+2)^2 , (x+2)^3),((x+3),(x+3)^2, (x+3)^3):}|` is expressed as a polynomial in x, then the term independent of x is :

Options

A. 0
B. 2
C. 12
D. 16

Answer

To solve the determinant

\[ D = \begin{vmatrix} x+1 & (x+1)^2 & (x+1)^3 \\ x+2 & (x+2)^2 & (x+2)^3 \\ x+3 & (x+3)^2 & (x+3)^3 \end{vmatrix} \]

we will expand it step by step.

### Step 1: Write the Determinant
We start with the determinant as given:

\[ D = \begin{vmatrix} x+1 & (x+1)^2 & (x+1)^3 \\ x+2 & (x+2)^2 & (x+2)^3 \\ x+3 & (x+3)^2 & (x+3)^3 \end{vmatrix} \]

### Step 2: Expand the Determinant
We can expand this determinant using the first row:

\[ D = (x+1) \begin{vmatrix} (x+2)^2 & (x+2)^3 \\ (x+3)^2 & (x+3)^3 \end{vmatrix} - (x+1)^2 \begin{vmatrix} x+2 & (x+2)^3 \\ x+3 & (x+3)^3 \end{vmatrix} + (x+1)^3 \begin{vmatrix} x+2 & (x+2)^2 \\ x+3 & (x+3)^2 \end{vmatrix} \]

### Step 3: Calculate the 2x2 Determinants
Now, we calculate each of the 2x2 determinants.

1. For the first determinant:

\[ \begin{vmatrix} (x+2)^2 & (x+2)^3 \\ (x+3)^2 & (x+3)^3 \end{vmatrix} = (x+2)^2(x+3)^3 - (x+2)^3(x+3)^2 \]

Factoring out common terms:

\[ = (x+2)^2(x+3)^2 \left( (x+3) - (x+2) \right) = (x+2)^2(x+3)^2 \]

2. For the second determinant:

\[ \begin{vmatrix} x+2 & (x+2)^3 \\ x+3 & (x+3)^3 \end{vmatrix} = (x+2)(x+3)^3 - (x+2)^3(x+3) \]

Factoring out common terms:

\[ = (x+2)(x+3) \left( (x+3)^2 - (x+2)^2 \right) = (x+2)(x+3)(2x + 5) \]

3. For the third determinant:

\[ \begin{vmatrix} x+2 & (x+2)^2 \\ x+3 & (x+3)^2 \end{vmatrix} = (x+2)(x+3)^2 - (x+2)^2(x+3) \]

Factoring out common terms:

\[ = (x+2)(x+3) \left( (x+3) - (x+2) \right) = (x+2)(x+3) \]

### Step 4: Substitute Back into the Determinant
Now substituting back into the expression for \(D\):

\[ D = (x+1)(x+2)^2(x+3)^2 - (x+1)^2(x+2)(x+3)(2x + 5) + (x+1)^3(x+2)(x+3) \]

### Step 5: Collect Terms
Now we can collect terms and simplify. The polynomial will be expressed in terms of powers of \(x\).

### Step 6: Identify the Constant Term
To find the term independent of \(x\), we can evaluate \(D\) at specific values of \(x\) that simplify calculations. A common choice is to evaluate at \(x = 0\):

\[ D(0) = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 4 & 8 \\ 3 & 9 & 27 \end{vmatrix} \]

Calculating this determinant gives:

\[ D(0) = 1 \cdot (4 \cdot 27 - 8 \cdot 9) - 1 \cdot (2 \cdot 27 - 8 \cdot 3) + 1 \cdot (2 \cdot 9 - 4 \cdot 3) \]

Calculating each term:

1. \(4 \cdot 27 - 8 \cdot 9 = 108 - 72 = 36\)
2. \(2 \cdot 27 - 8 \cdot 3 = 54 - 24 = 30\)
3. \(2 \cdot 9 - 4 \cdot 3 = 18 - 12 = 6\)

Putting it all together:

\[ D(0) = 1 \cdot 36 - 1 \cdot 30 + 1 \cdot 6 = 36 - 30 + 6 = 12 \]

### Final Answer
Thus, the term independent of \(x\) is:

\[ \boxed{12} \]