If s=(a+b+c),then value of `|{:(s+c,a,b),(c,s+a,b),(c,a,s+b):}|`is

Options

A. `2s^2`
B. `2s^3`
C. `s^3`
D. `3s^3`

Answer

To solve the determinant \( | \begin{pmatrix} s+c & a & b \\ c & s+a & b \\ c & a & s+b \end{pmatrix} | \) where \( s = a + b + c \), we can follow these steps:

### Step 1: Substitute \( s \)
First, we substitute \( s \) with \( a + b + c \) in the determinant.

\[ | \begin{pmatrix} (a+b+c) + c & a & b \\ c & (a+b+c) + a & b \\ c & a & (a+b+c) + b \end{pmatrix} | \]

This simplifies to:

\[ | \begin{pmatrix} a+b+2c & a & b \\ c & a+b+c & b \\ c & a & a+b+c \end{pmatrix} | \]

### Step 2: Perform Row Operations
We will perform row operations to simplify the determinant. Let's subtract the second row from the first row:

\[ R_1 \rightarrow R_1 - R_2 \]

This gives us:

\[ | \begin{pmatrix} (a+b+2c) - c & a - (a+b+c) & b - b \\ c & a+b+c & b \\ c & a & a+b+c \end{pmatrix} | \]

Calculating the first row:

- First element: \( (a+b+2c) - c = a + b + c \)
- Second element: \( a - (a+b+c) = -b - c \)
- Third element: \( b - b = 0 \)

So, the determinant now looks like:

\[ | \begin{pmatrix} a+b+c & -b-c & 0 \\ c & a+b+c & b \\ c & a & a+b+c \end{pmatrix} | \]

### Step 3: Factor Out Common Terms
Next, we can factor out \( c \) from the second and third rows:

\[ | \begin{pmatrix} a+b+c & -b-c & 0 \\ c & a+b+c & b \\ c & a & a+b+c \end{pmatrix} | \]

### Step 4: Further Row Operations
Now, let's perform another row operation: \( R_2 \rightarrow R_2 - R_3 \):

\[ | \begin{pmatrix} a+b+c & -b-c & 0 \\ c - c & (a+b+c) - a & b - (a+b+c) \\ c & a & a+b+c \end{pmatrix} | \]

This results in:

\[ | \begin{pmatrix} a+b+c & -b-c & 0 \\ 0 & b+c & -a \\ c & a & a+b+c \end{pmatrix} | \]

### Step 5: Expand the Determinant
Now we can expand this determinant:

\[ = (a+b+c) \cdot | \begin{pmatrix} b+c & -a \\ a & a+b+c \end{pmatrix} | \]

Calculating the 2x2 determinant:

\[ = (a+b+c) \cdot ((b+c)(a+b+c) - (-a)(a)) = (a+b+c) \cdot (ab + ac + bc + b^2 + c^2 + a^2) \]

### Step 6: Substitute Back \( s \)
Since \( s = a + b + c \):

\[ = s \cdot (s^2) = s^3 \]

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

\[ \boxed{2s^3} \]