If `|(a^2,b^2,c^2),((a+1)^2 ,(b+1)^2,(c+1)^2),((a-1)^2 ,(b-1)^2,(c-1)^2)| =k(a-b)(b-c)(c-a)` then the value of k is a. 4 b. -2 c.-4 d. 2

Explanation

To solve the problem, we need to evaluate the determinant of the following matrix:

\[ D = \begin{vmatrix} a^2 & b^2 & c^2 \\ (a+1)^2 & (b+1)^2 & (c+1)^2 \\ (a-1)^2 & (b-1)^2 & (c-1)^2 \end{vmatrix} \]

We are given that this determinant equals \( k(a-b)(b-c)(c-a) \) and we need to find the value of \( k \).

### Step 1: Write the matrix

The matrix is:

\[ D = \begin{vmatrix} a^2 & b^2 & c^2 \\ (a+1)^2 & (b+1)^2 & (c+1)^2 \\ (a-1)^2 & (b-1)^2 & (c-1)^2 \end{vmatrix} \]

### Step 2: Apply row transformations

We will perform a row operation to simplify the determinant. Specifically, we will subtract the third row from the second row:

\[ R_2 \rightarrow R_2 - R_3 \]

This gives us:

\[ D = \begin{vmatrix} a^2 & b^2 & c^2 \\ (a+1)^2 - (a-1)^2 & (b+1)^2 - (b-1)^2 & (c+1)^2 - (c-1)^2 \\ (a-1)^2 & (b-1)^2 & (c-1)^2 \end{vmatrix} \]

Calculating the elements of the new second row:

\[ (a+1)^2 - (a-1)^2 = 4a, \quad (b+1)^2 - (b-1)^2 = 4b, \quad (c+1)^2 - (c-1)^2 = 4c \]

Thus, we have:

\[ D = \begin{vmatrix} a^2 & b^2 & c^2 \\ 4a & 4b & 4c \\ (a-1)^2 & (b-1)^2 & (c-1)^2 \end{vmatrix} \]

### Step 3: Factor out constants

We can factor out 4 from the second row:

\[ D = 4 \begin{vmatrix} a^2 & b^2 & c^2 \\ a & b & c \\ (a-1)^2 & (b-1)^2 & (c-1)^2 \end{vmatrix} \]

### Step 4: Apply another row transformation

Now, we will perform another row operation on the third row:

\[ R_3 \rightarrow R_3 - R_1 \]

This gives us:

\[ D = 4 \begin{vmatrix} a^2 & b^2 & c^2 \\ a & b & c \\ (a-1)^2 - a^2 & (b-1)^2 - b^2 & (c-1)^2 - c^2 \end{vmatrix} \]

Calculating the elements of the new third row:

\[ (a-1)^2 - a^2 = -2a + 1, \quad (b-1)^2 - b^2 = -2b + 1, \quad (c-1)^2 - c^2 = -2c + 1 \]

Thus, we have:

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

### Step 5: Evaluate the determinant

Now we can evaluate the determinant. The determinant of this matrix can be computed using the formula for determinants of 3x3 matrices.

After evaluating, we find that:

\[ D = -4(a-b)(b-c)(c-a) \]

### Step 6: Compare with the given expression

We have:

\[ D = k(a-b)(b-c)(c-a) \]

Comparing both sides, we find:

\[ k = -4 \]

### Conclusion

Thus, the value of \( k \) is:

\[ \boxed{-4} \]