The value of the determinant
`|(1+a,1,1),(1,1+a,1),(1,1,1+a)|` is

Options

A. `a^3(1-2/a)`
B. `a^3(1+3/a)`
C. `a^3(1-3/a)`
D. `a^3(1+2/a)`

Answer

To find the value of the determinant

\[ D = \begin{vmatrix} 1 + a & 1 & 1 \\ 1 & 1 + a & 1 \\ 1 & 1 & 1 + a \end{vmatrix} \]

we will expand it along the first row.

### Step 1: Expand the Determinant

Using the first row for expansion, we have:

\[ D = (1 + a) \begin{vmatrix} 1 + a & 1 \\ 1 & 1 + a \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & 1 + a \end{vmatrix} + 1 \begin{vmatrix} 1 & 1 + a \\ 1 & 1 \end{vmatrix} \]

### Step 2: Calculate the 2x2 Determinants

Now, we calculate each of the 2x2 determinants:

1. For the first determinant:

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

2. For the second determinant:

\[ \begin{vmatrix} 1 & 1 \\ 1 & 1 + a \end{vmatrix} = (1)(1 + a) - (1)(1) = 1 + a - 1 = a \]

3. For the third determinant:

\[ \begin{vmatrix} 1 & 1 + a \\ 1 & 1 \end{vmatrix} = (1)(1) - (1)(1 + a) = 1 - (1 + a) = -a \]

### Step 3: Substitute Back into the Determinant

Now substituting these values back into the expression for \(D\):

\[ D = (1 + a)(a^2 + 2a) - 1(a) + 1(-a) \]

This simplifies to:

\[ D = (1 + a)(a^2 + 2a) - a - a \]

\[ D = (1 + a)(a^2 + 2a) - 2a \]

### Step 4: Expand the Expression

Now, we expand \((1 + a)(a^2 + 2a)\):

\[ D = a^2 + 2a + a^3 + 2a^2 - 2a \]

Combining like terms:

\[ D = a^3 + 3a^2 \]

### Final Result

Thus, the value of the determinant is:

\[ D = a^3 + 3a^2 \]