If `Delta_1=|{:(1,1,1),(a,b,c),(a^2,b^2,c^2):}|` and `Delta_2=|{:(1,bc,a),(1,ac,b),(1,ab,c):}|` then : -

Options

A. `Delta_1=Delta_2`
B. `Delta_1=2Delta_2`
C. `Delta_1+Delta_2=0`
D. `Delta_1+2Delta_2=0`

Answer

To solve the problem, we need to evaluate the determinants \( \Delta_1 \) and \( \Delta_2 \) and then check the relationships between them.

### Step 1: Evaluate \( \Delta_1 \)

Given:
\[ \Delta_1 = \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} \]

We can use the properties of determinants to simplify this. We will expand the determinant along the first row.

\[ \Delta_1 = 1 \cdot \begin{vmatrix} b & c \\ b^2 & c^2 \end{vmatrix} - 1 \cdot \begin{vmatrix} a & c \\ a^2 & c^2 \end{vmatrix} + 1 \cdot \begin{vmatrix} a & b \\ a^2 & b^2 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} b & c \\ b^2 & c^2 \end{vmatrix} = bc^2 - b^2c \)
2. \( \begin{vmatrix} a & c \\ a^2 & c^2 \end{vmatrix} = ac^2 - a^2c \)
3. \( \begin{vmatrix} a & b \\ a^2 & b^2 \end{vmatrix} = ab^2 - a^2b \)

Putting it all together:
\[ \Delta_1 = (bc^2 - b^2c) - (ac^2 - a^2c) + (ab^2 - a^2b) \]
\[ = bc^2 - b^2c - ac^2 + a^2c + ab^2 - a^2b \]

### Step 2: Evaluate \( \Delta_2 \)

Given:
\[ \Delta_2 = \begin{vmatrix} 1 & bc & a \\ 1 & ac & b \\ 1 & ab & c \end{vmatrix} \]

Again, we will expand this determinant along the first row.

\[ \Delta_2 = 1 \cdot \begin{vmatrix} ac & b \\ ab & c \end{vmatrix} - 1 \cdot \begin{vmatrix} bc & a \\ ab & c \end{vmatrix} + 1 \cdot \begin{vmatrix} bc & a \\ ac & b \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} ac & b \\ ab & c \end{vmatrix} = ac^2 - ab^2 \)
2. \( \begin{vmatrix} bc & a \\ ab & c \end{vmatrix} = bca - a^2b \)
3. \( \begin{vmatrix} bc & a \\ ac & b \end{vmatrix} = b^2c - a^2c \)

Putting it all together:
\[ \Delta_2 = (ac^2 - ab^2) - (bca - a^2b) + (b^2c - a^2c) \]
\[ = ac^2 - ab^2 - bca + a^2b + b^2c - a^2c \]

### Step 3: Check if \( \Delta_1 + \Delta_2 = 0 \)

Now we need to find \( \Delta_1 + \Delta_2 \) and check if it equals zero.

From our previous calculations:
\[ \Delta_1 + \Delta_2 = (bc^2 - b^2c - ac^2 + a^2c + ab^2 - a^2b) + (ac^2 - ab^2 - bca + a^2b + b^2c - a^2c) \]

Combining like terms:
- The \( ac^2 \) and \( -ac^2 \) cancel.
- The \( ab^2 \) and \( -ab^2 \) cancel.
- The \( -b^2c \) and \( b^2c \) cancel.
- The \( -bca \) remains.
- The \( a^2c \) and \( -a^2c \) cancel.
- The \( a^2b \) and \( -a^2b \) cancel.

Thus, we find:
\[ \Delta_1 + \Delta_2 = -bca + bca = 0 \]

### Conclusion

Therefore, we conclude that:
\[ \Delta_1 + \Delta_2 = 0 \]