Value of the `D=|(1/a,bc,a^(3)),(1/b,ca,b^(3)),(1/c,ab,c^(3))|` is

Options

A. 0
B. `(a^(3)-1)(a^(6)-1)(a^(9)-1)`
C. `(a^(3)+1)(a^(6)+1)(a^(9)+1)`
D. `a^(15)-1`

Answer

To find the value of the determinant \( D = \begin{vmatrix} \frac{1}{a} & bc & a^3 \\ \frac{1}{b} & ca & b^3 \\ \frac{1}{c} & ab & c^3 \end{vmatrix} \), we will follow these steps:

### Step 1: Factor out common terms
We can factor out \( \frac{1}{abc} \) from the first column. This gives us:

\[ D = \frac{1}{abc} \begin{vmatrix} 1 & bc & a^3 \\ 1 & ca & b^3 \\ 1 & ab & c^3 \end{vmatrix} \]

**Hint:** Look for common factors in the rows or columns to simplify the determinant.

### Step 2: Perform row operations
Next, we can perform row operations to simplify the determinant. We can subtract the first row from the second and third rows:

\[ D = \frac{1}{abc} \begin{vmatrix} 1 & bc & a^3 \\ 0 & ca - bc & b^3 - a^3 \\ 0 & ab - bc & c^3 - a^3 \end{vmatrix} \]

**Hint:** Row operations can help eliminate variables and simplify the determinant.

### Step 3: Factor out common terms from the second and third rows
Now, we can factor out \( (ca - bc) \) from the second row and \( (ab - bc) \) from the third row:

\[ D = \frac{1}{abc} \cdot (ca - bc) \cdot (ab - bc) \begin{vmatrix} 1 & bc & a^3 \\ 0 & 1 & \frac{b^3 - a^3}{ca - bc} \\ 0 & 1 & \frac{c^3 - a^3}{ab - bc} \end{vmatrix} \]

**Hint:** Look for patterns in the rows or columns that can be factored out.

### Step 4: Analyze the determinant
Now, we can see that the second and third rows are identical if \( b^3 - a^3 \) and \( c^3 - a^3 \) are equal. This means that the determinant will equal zero if two rows are identical:

\[ D = 0 \]

**Hint:** If two rows or columns of a determinant are identical, the determinant is zero.

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

\[ \boxed{0} \]