If the polynomial
`f(x) = |{:((1+x)^(a), (2+x)^(b),1),(1, (1 + x)^(a), (2 + x)^(b)),((2+x)^(b) , 1, (1+x)^(a)) :}|`
then the constant term of f(x) is

Options

A. 2 - 3.`2^(b) + 2^(3b)`
B. `2 + 3.2^(b) + 2^(3b)`
C. `2 + 3.2^(b)-2^(3b)`
D. `2- 3.2^(b)-2^(3b)`

Answer

To find the constant term of the polynomial given by the determinant:

\[ f(x) = \begin{vmatrix} (1+x)^{a} & (2+x)^{b} & 1 \\ 1 & (1+x)^{a} & (2+x)^{b} \\ (2+x)^{b} & 1 & (1+x)^{a} \end{vmatrix} \]

we will evaluate the determinant by substituting \(x = 0\).

### Step 1: Substitute \(x = 0\) into the determinant

When we substitute \(x = 0\), we get:

\[ f(0) = \begin{vmatrix} (1+0)^{a} & (2+0)^{b} & 1 \\ 1 & (1+0)^{a} & (2+0)^{b} \\ (2+0)^{b} & 1 & (1+0)^{a} \end{vmatrix} \]

This simplifies to:

\[ f(0) = \begin{vmatrix} 1^{a} & 2^{b} & 1 \\ 1 & 1^{a} & 2^{b} \\ 2^{b} & 1 & 1^{a} \end{vmatrix} \]

Since \(1^{a} = 1\) for any \(a\), we can further simplify this to:

\[ f(0) = \begin{vmatrix} 1 & 2^{b} & 1 \\ 1 & 1 & 2^{b} \\ 2^{b} & 1 & 1 \end{vmatrix} \]

### Step 2: Calculate the determinant

Now we will calculate the determinant using the first row:

\[ f(0) = 1 \cdot \begin{vmatrix} 1 & 2^{b} \\ 1 & 1 \end{vmatrix} - 2^{b} \cdot \begin{vmatrix} 1 & 2^{b} \\ 2^{b} & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 1 \\ 2^{b} & 1 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \(\begin{vmatrix} 1 & 2^{b} \\ 1 & 1 \end{vmatrix} = (1)(1) - (1)(2^{b}) = 1 - 2^{b}\)

2. \(\begin{vmatrix} 1 & 2^{b} \\ 2^{b} & 1 \end{vmatrix} = (1)(1) - (2^{b})(2^{b}) = 1 - 2^{2b}\)

3. \(\begin{vmatrix} 1 & 1 \\ 2^{b} & 1 \end{vmatrix} = (1)(1) - (1)(2^{b}) = 1 - 2^{b}\)

Substituting these back into the determinant calculation:

\[ f(0) = 1(1 - 2^{b}) - 2^{b}(1 - 2^{2b}) + 1(1 - 2^{b}) \]

### Step 3: Simplify the expression

Now we simplify:

\[ f(0) = (1 - 2^{b}) - 2^{b}(1 - 2^{2b}) + (1 - 2^{b}) \]

Combining like terms:

\[ f(0) = 2(1 - 2^{b}) - 2^{b}(1 - 2^{2b}) \]

Expanding the second term:

\[ = 2 - 2 \cdot 2^{b} - 2^{b} + 2^{3b} \]

This simplifies to:

\[ = 2 - 3 \cdot 2^{b} + 2^{3b} \]

### Final Answer

Thus, the constant term of \(f(x)\) is:

\[ \boxed{2 - 3 \cdot 2^{b} + 2^{3b}} \]