`|(1/a,a^2,bc),(1/b,b^2,ca),(1/c,c^2,ab)|=........`

Answer

To solve the determinant \( D = \begin{vmatrix} \frac{1}{a} & a^2 & bc \\ \frac{1}{b} & b^2 & ca \\ \frac{1}{c} & c^2 & ab \end{vmatrix} \), we will follow these steps:

### Step 1: Factor out common terms from each row
We can factor out \( a \) from the first row, \( b \) from the second row, and \( c \) from the third row. This gives us:

\[ D = \begin{vmatrix} \frac{1}{a} & a^2 & bc \\ \frac{1}{b} & b^2 & ca \\ \frac{1}{c} & c^2 & ab \end{vmatrix} = \frac{1}{a} \cdot \frac{1}{b} \cdot \frac{1}{c} \begin{vmatrix} 1 & a^3 & abc \\ 1 & b^3 & abc \\ 1 & c^3 & abc \end{vmatrix} \]

### Step 2: Simplifying the determinant
Now we can simplify the determinant:

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

### Step 3: Subtract the first row from the second and third rows
Next, we can perform row operations to simplify the determinant further. Subtract the first row from the second and third rows:

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

### Step 4: Evaluate the determinant
Notice that the last two rows have a zero in the last column. Thus, the determinant evaluates to zero:

\[ D = \frac{1}{abc} \cdot 0 = 0 \]

### Final Answer
The value of the determinant is:

\[ \boxed{0} \]