The determinat `Delta=|(b^2-ab,b-c,bc-ac),(ab-a^2,a-b,b^2-ab),(bc-ac,c-a,ab-a^2)|` equals

Options

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

Explanation

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

### Step 1: Factor out common terms from the columns
Observe that we can factor out common terms from the columns of the determinant.

1. From the first column, we can factor out \( b - a \).
2. From the second column, we can factor out \( b - a \).
3. From the third column, we can factor out \( c - a \).

This gives us:
\[ \Delta = (b - a)^2 \cdot (c - a) \begin{vmatrix} b & b - c & c \\ a & a - b & b \\ 0 & c - a & a \end{vmatrix} \]

### Step 2: Simplify the determinant
Now we can simplify the determinant further. Notice that we can perform column operations to simplify it.

We can replace \( C_1 \) with \( C_1 - C_3 \):
\[ \Delta = (b - a)^2 (c - a) \begin{vmatrix} b - c & b - c & c \\ a - b & a - b & b \\ 0 & c - a & a \end{vmatrix} \]

### Step 3: Identify equal columns
From the determinant, we can see that the first and second columns are now equal:
\[ \Delta = (b - a)^2 (c - a) \begin{vmatrix} b - c & b - c & c \\ a - b & a - b & b \\ 0 & c - a & a \end{vmatrix} \]

### Step 4: Conclude the determinant value
Since two columns of the determinant are equal, the value of the determinant is zero:
\[ \Delta = 0 \]

### Final Answer
Thus, the value of the determinant \( \Delta \) is:
\[ \Delta = 0 \]

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