Suppose `a,b epsilon R` and `|(x,a,b),(a,x,b),(b,b,x)|-4k(x-a)(x^(2)+ax-2b^(2))=0` then a value of k is ________

Answer

To solve the problem, we need to evaluate the determinant and set it equal to the given expression. Let's go through the steps systematically.

### Step 1: Write the Determinant
We are given the determinant:
\[ D = \begin{vmatrix} x & a & b \\ a & x & b \\ b & b & x \end{vmatrix} \]

### Step 2: Expand the Determinant
We can expand the determinant using the first row:
\[ D = x \begin{vmatrix} x & b \\ b & x \end{vmatrix} - a \begin{vmatrix} a & b \\ b & x \end{vmatrix} + b \begin{vmatrix} a & x \\ b & b \end{vmatrix} \]

### Step 3: Calculate the 2x2 Determinants
Now, we calculate each of the 2x2 determinants:
1. \(\begin{vmatrix} x & b \\ b & x \end{vmatrix} = x^2 - b^2\)
2. \(\begin{vmatrix} a & b \\ b & x \end{vmatrix} = ax - b^2\)
3. \(\begin{vmatrix} a & x \\ b & b \end{vmatrix} = ab - bx\)

### Step 4: Substitute Back into the Determinant
Substituting these back into the expression for \(D\):
\[ D = x(x^2 - b^2) - a(ax - b^2) + b(ab - bx) \]
\[ = x^3 - xb^2 - a^2x + ab^2 + ab - b^2x \]
\[ = x^3 - (a^2 + 2b^2)x + ab^2 + ab \]

### Step 5: Factor the Determinant
We need to factor the expression \(D\) in terms of \(x-a\):
\[ D = (x-a)(x^2 + ax - 2b^2) \]

### Step 6: Set the Determinant Equal to the Given Expression
According to the problem, we have:
\[ |D| - 4k(x-a)(x^2 + ax - 2b^2) = 0 \]
This implies:
\[ D = 4k(x-a)(x^2 + ax - 2b^2) \]

### Step 7: Compare Coefficients
From our factorization, we can see that:
\[ 1 = 4k \]
Thus, solving for \(k\):
\[ k = \frac{1}{4} \]

### Final Answer
The value of \(k\) is:
\[ \boxed{\frac{1}{4}} \]