The value of the determinant `|(a+b,a+2b,a+3b),(a+2b,a+3b,a+4b),(a+4b,a+5b,a+6b)|` is

Options

A. 4abc
B. 0
C. `a^2+b^2`
D. none

Answer

To find the value of the determinant

\[ D = \begin{vmatrix} a+b & a+2b & a+3b \\ a+2b & a+3b & a+4b \\ a+4b & a+5b & a+6b \end{vmatrix} \]

we can use row and column operations to simplify the determinant.

### Step 1: Apply Column Operations
Let's perform the following column operations:
- \( C_1 \leftarrow C_1 - C_3 \)
- \( C_2 \leftarrow C_2 - C_3 \)

This gives us:

\[ D = \begin{vmatrix} (a+b) - (a+3b) & (a+2b) - (a+3b) & a+3b \\ (a+2b) - (a+4b) & (a+3b) - (a+4b) & a+4b \\ (a+4b) - (a+6b) & (a+5b) - (a+6b) & a+6b \end{vmatrix} \]

Calculating each entry:

- First row:
- First element: \( (a+b) - (a+3b) = -2b \)
- Second element: \( (a+2b) - (a+3b) = -b \)

- Second row:
- First element: \( (a+2b) - (a+4b) = -2b \)
- Second element: \( (a+3b) - (a+4b) = -b \)

- Third row:
- First element: \( (a+4b) - (a+6b) = -2b \)
- Second element: \( (a+5b) - (a+6b) = -b \)

So, we have:

\[ D = \begin{vmatrix} -2b & -b & a+3b \\ -2b & -b & a+4b \\ -2b & -b & a+6b \end{vmatrix} \]

### Step 2: Factor Out Common Terms
Notice that the first two columns have a common factor of \(-b\) and the first column has a common factor of \(-2b\). We can factor these out:

\[ D = (-b)(-b)(-2b) \begin{vmatrix} 1 & 1 & a+3b \\ 1 & 1 & a+4b \\ 1 & 1 & a+6b \end{vmatrix} \]

This simplifies to:

\[ D = 2b^3 \begin{vmatrix} 1 & 1 & a+3b \\ 1 & 1 & a+4b \\ 1 & 1 & a+6b \end{vmatrix} \]

### Step 3: Evaluate the 3x3 Determinant
Now, we can evaluate the determinant:

\[ \begin{vmatrix} 1 & 1 & a+3b \\ 1 & 1 & a+4b \\ 1 & 1 & a+6b \end{vmatrix} \]

Notice that the first two columns are identical, which means the determinant is zero:

\[ D = 2b^3 \cdot 0 = 0 \]

### Final Answer
Thus, the value of the determinant is:

\[ \boxed{0} \]