`|(1,1,1),(a,b,c),(a^3,b^3,c^3)|`=

Options

A. `a^3+b^3+c^3 -3abc `
B. `a^3+b^3+c^3 +3abc `
C. `(a+b+c)(a-b)(b-c)(c-a)`
D. none of these

Answer

To solve the determinant \( D = \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{vmatrix} \), we will follow a systematic approach.

### Step-by-Step Solution:

1. **Write the Determinant:**
\[ D = \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{vmatrix} \]

2. **Perform Column Operations:**
We can simplify the determinant by performing column operations. Let's perform the following operations:
- \( C_1 \leftarrow C_1 - C_2 \)
- \( C_2 \leftarrow C_2 - C_3 \)

After performing these operations, the determinant becomes:
\[ D = \begin{vmatrix} 0 & 0 & 1 \\ a-b & b-c & c \\ a^3-b^3 & b^3-c^3 & c^3 \end{vmatrix} \]

3. **Evaluate the Determinant:**
Since the first two columns have become zero, we can expand the determinant along the first row:
\[ D = 1 \cdot \begin{vmatrix} a-b & b-c \\ a^3-b^3 & b^3-c^3 \end{vmatrix} \]

4. **Use the Difference of Cubes Formula:**
Recall the identity for the difference of cubes:
\[ x^3 - y^3 = (x-y)(x^2 + xy + y^2) \]
Thus, we can express \( a^3 - b^3 \) and \( b^3 - c^3 \) as:
\[ a^3 - b^3 = (a-b)(a^2 + ab + b^2) \]
\[ b^3 - c^3 = (b-c)(b^2 + bc + c^2) \]

5. **Substituting Back:**
Substitute these into the determinant:
\[ D = (a-b)(b-c) \begin{vmatrix} 1 & 1 \\ a^2 + ab + b^2 & b^2 + bc + c^2 \end{vmatrix} \]

6. **Calculate the Remaining Determinant:**
The remaining determinant is:
\[ D = (a-b)(b-c) \left( (1)(b^2 + bc + c^2) - (1)(a^2 + ab + b^2) \right) \]
Simplifying this gives:
\[ D = (a-b)(b-c)(b^2 + bc + c^2 - a^2 - ab - b^2) \]
\[ D = (a-b)(b-c)(c^2 - a^2 - ab) \]

7. **Final Result:**
The final expression can be factored further:
\[ D = (a-b)(b-c)(c-a)(a+b+c) \]

### Final Answer:
\[ D = (a-b)(b-c)(c-a)(a+b+c) \]