Let `Delta=|(a,a^(2),0),(1,2a+b,(a+b)^(2)),(0,1,2a+3b)|` then

Options

A. `a + b` is a factor of `Delta`
B. `a + 2b` is a factor of `Delta`
C. `2a+3b` is a factor of `Delta`
D. `a^(2)` is a factor of `Delta`

Answer

To solve the determinant \( \Delta = \begin{vmatrix} a & a^2 & 0 \\ 1 & 2a + b & (a + b)^2 \\ 0 & 1 & 2a + 3b \end{vmatrix} \), we will expand it along the first row.

### Step 1: Write the determinant
We start with the determinant as given:
\[ \Delta = \begin{vmatrix} a & a^2 & 0 \\ 1 & 2a + b & (a + b)^2 \\ 0 & 1 & 2a + 3b \end{vmatrix} \]

### Step 2: Expand along the first row
Using the cofactor expansion along the first row, we have:
\[ \Delta = a \cdot \begin{vmatrix} 2a + b & (a + b)^2 \\ 1 & 2a + 3b \end{vmatrix} - a^2 \cdot \begin{vmatrix} 1 & (a + b)^2 \\ 0 & 2a + 3b \end{vmatrix} + 0 \]

### Step 3: Calculate the first minor
Calculate the first minor:
\[ \begin{vmatrix} 2a + b & (a + b)^2 \\ 1 & 2a + 3b \end{vmatrix} = (2a + b)(2a + 3b) - (a + b)^2 \]
Expanding this:
\[ = (4a^2 + 6ab + 2ab + 3b^2) - (a^2 + 2ab + b^2) = 4a^2 + 8ab + 3b^2 - a^2 - 2ab - b^2 \]
\[ = 3a^2 + 6ab + 2b^2 \]

### Step 4: Calculate the second minor
Calculate the second minor:
\[ \begin{vmatrix} 1 & (a + b)^2 \\ 0 & 2a + 3b \end{vmatrix} = 1 \cdot (2a + 3b) - 0 = 2a + 3b \]

### Step 5: Substitute back into the determinant
Now substitute these minors back into the expression for \(\Delta\):
\[ \Delta = a(3a^2 + 6ab + 2b^2) - a^2(2a + 3b) \]
Expanding this gives:
\[ = 3a^3 + 6a^2b + 2ab^2 - (2a^3 + 3a^2b) \]
\[ = (3a^3 - 2a^3) + (6a^2b - 3a^2b) + 2ab^2 = a^3 + 3a^2b + 2ab^2 \]

### Step 6: Factor the result
We can factor out \(a\) from the expression:
\[ \Delta = a(a^2 + 3ab + 2b^2) \]
Now we can factor the quadratic:
\[ a^2 + 3ab + 2b^2 = (a + b)(a + 2b) \]
Thus, we have:
\[ \Delta = a(a + b)(a + 2b) \]

### Final Result
Therefore, the final expression for the determinant is:
\[ \Delta = a(a + b)(a + 2b) \]