Let a+b+c =s and `|{:(s+c,,a,,b),(c,,s+a,,b),(c,,a,,s+b):}|=432` then the value of s is `"____"`

Explanation

To solve the problem step by step, we start with the given determinant and the relationship \( a + b + c = s \).

### Step 1: Write the determinant
We have the determinant:
\[ D = \begin{vmatrix} s+c & a & b \\ c & s+a & b \\ c & a & s+b \end{vmatrix} \]
and we know that \( D = 432 \).

### Step 2: Simplify the determinant
We can apply the column operation \( C_1 \to C_1 + C_2 + C_3 \):
\[ D = \begin{vmatrix} (s+c) + a + b & a & b \\ c + (s+a) + b & s+a & b \\ c + a + (s+b) & a & s+b \end{vmatrix} \]
This simplifies to:
\[ D = \begin{vmatrix} s + (a+b+c) & a & b \\ s + (a+b+c) & s+a & b \\ s + (a+b+c) & a & s+b \end{vmatrix} \]
Since \( a + b + c = s \), we can replace \( a + b + c \) with \( s \):
\[ D = \begin{vmatrix} 2s & a & b \\ 2s & s+a & b \\ 2s & a & s+b \end{vmatrix} \]

### Step 3: Factor out common terms
We can factor out \( 2s \) from the first column:
\[ D = 2s \begin{vmatrix} 1 & a & b \\ 1 & s+a & b \\ 1 & a & s+b \end{vmatrix} \]

### Step 4: Perform row operations
Next, we can perform row operations to simplify the determinant:
- \( R_2 \to R_2 - R_1 \)
- \( R_3 \to R_3 - R_1 \)

This gives us:
\[ D = 2s \begin{vmatrix} 1 & a & b \\ 0 & (s+a - a) & (b - b) \\ 0 & (a - a) & (s+b - b) \end{vmatrix} \]
Which simplifies to:
\[ D = 2s \begin{vmatrix} 1 & a & b \\ 0 & s & 0 \\ 0 & 0 & s \end{vmatrix} \]

### Step 5: Calculate the determinant
The determinant of the above matrix is:
\[ D = 2s \cdot s \cdot s = 2s^3 \]

### Step 6: Set the determinant equal to 432
Now, we set the expression equal to 432:
\[ 2s^3 = 432 \]

### Step 7: Solve for \( s \)
Dividing both sides by 2:
\[ s^3 = 216 \]
Taking the cube root:
\[ s = \sqrt[3]{216} = 6 \]

### Final Answer
Thus, the value of \( s \) is:
\[ \boxed{6} \]