If `f(x)=|((1+x)^(17),(a+x)^(19),(1+x)^(23)),((a+x)^(23),(a+x)^(29),(1+x)^(34)),((1+x)^(41),(1+x)^(43),(1+x)^(47))|`
`=A+Bx+Cx^(2)+……….` then `A` is equal to………….

Explanation

To solve the problem, we need to evaluate the determinant given by the function \( f(x) \) and find the value of \( A \) in the expression \( f(x) = A + Bx + Cx^2 + \ldots \).

The determinant is defined as follows:

\[ f(x) = \begin{vmatrix} (1+x)^{17} & (a+x)^{19} & (1+x)^{23} \\ (a+x)^{23} & (a+x)^{29} & (1+x)^{34} \\ (1+x)^{41} & (1+x)^{43} & (1+x)^{47} \end{vmatrix} \]

### Step 1: Substitute \( x = 0 \)

To find \( A \), we can evaluate \( f(0) \):

\[ f(0) = \begin{vmatrix} (1+0)^{17} & (a+0)^{19} & (1+0)^{23} \\ (a+0)^{23} & (a+0)^{29} & (1+0)^{34} \\ (1+0)^{41} & (1+0)^{43} & (1+0)^{47} \end{vmatrix} \]

This simplifies to:

\[ f(0) = \begin{vmatrix} 1 & a^{19} & 1 \\ a^{23} & a^{29} & 1 \\ 1 & 1 & 1 \end{vmatrix} \]

### Step 2: Calculate the Determinant

Now, we compute the determinant:

Using the properties of determinants, we can simplify this determinant. Notice that the first and third columns are identical in terms of their first and last entries, which will lead to a simplification.

Calculating the determinant, we can perform row operations or use the determinant formula directly. However, we can also see that the first and third rows are identical, which means:

\[ f(0) = 0 \]

### Step 3: Conclusion

Since \( f(0) = A \), we have:

\[ A = 0 \]

Thus, the value of \( A \) is:

\[ \boxed{0} \]