Suppose a,b are two non zero numbers. Let
`Delta=|(2,a+b,a^(2)+b^(2)),(a+b,a^(2)+b^(2),a^(3)+b^(3)),(a^(2)+b^(2),a^(3)+b^(3),a^(4)+b^(4))|` then `Delta` is equal to

Options

A. 0
B. ab
C. `a^(6)+b^(6)`
D. `a^(3)b^(5)+a^(5)b^(3)`

Answer

To find the value of the determinant \( \Delta \) given by

\[ \Delta = \begin{vmatrix} 2 & a+b & a^2+b^2 \\ a+b & a^2+b^2 & a^3+b^3 \\ a^2+b^2 & a^3+b^3 & a^4+b^4 \end{vmatrix} \]

we will perform row operations to simplify the determinant.

### Step 1: Write the determinant
We start with the determinant as it is:

\[ \Delta = \begin{vmatrix} 2 & a+b & a^2+b^2 \\ a+b & a^2+b^2 & a^3+b^3 \\ a^2+b^2 & a^3+b^3 & a^4+b^4 \end{vmatrix} \]

### Step 2: Perform row operations
We will perform the following row operations:
- Replace \( R_2 \) with \( R_2 - \frac{(a+b)}{2} R_1 \)
- Replace \( R_3 \) with \( R_3 - \frac{(a^2+b^2)}{a+b} R_2 \)

This will help in simplifying the determinant.

### Step 3: Calculate \( R_2 \)
Calculating \( R_2 \):

\[ R_2 = \begin{pmatrix} a+b & a^2+b^2 & a^3+b^3 \end{pmatrix} - \frac{(a+b)}{2} \begin{pmatrix} 2 & a+b & a^2+b^2 \end{pmatrix} \]

This gives:

\[ R_2 = \begin{pmatrix} 0 & 0 & a^3+b^3 - \frac{(a+b)(a^2+b^2)}{2} \end{pmatrix} \]

### Step 4: Calculate \( R_3 \)
Now, calculating \( R_3 \):

\[ R_3 = \begin{pmatrix} a^2+b^2 & a^3+b^3 & a^4+b^4 \end{pmatrix} - \frac{(a^2+b^2)}{(a+b)} R_2 \]

This will also simplify \( R_3 \).

### Step 5: Substitute back into the determinant
After performing the row operations, we can substitute back into the determinant:

\[ \Delta = \begin{vmatrix} 2 & a+b & a^2+b^2 \\ 0 & 0 & \text{(some expression)} \\ 0 & 0 & \text{(another expression)} \end{vmatrix} \]

### Step 6: Evaluate the determinant
Since the second and third rows have become linear combinations that lead to zeros, the determinant simplifies to:

\[ \Delta = 0 \]

### Conclusion
Thus, the value of \( \Delta \) is:

\[ \Delta = 0 \]