`ab(a-b)+bc(b-c)+ca(c-a)` is equal to

Options

A. `(a+b)(b-c)(c-a)`
B. `(b-a)(b-c)(c-a)`
C. `(a-b)(b-c)(c-a)`
D. `(a-b)(b+c)(c-a)`

Answer

To solve the expression \( ab(a-b) + bc(b-c) + ca(c-a) \), we can follow these steps:

### Step 1: Expand the expression
We start by expanding each term in the expression:

1. Expand \( ab(a-b) \):
\[ ab(a-b) = a^2b - ab^2 \]

2. Expand \( bc(b-c) \):
\[ bc(b-c) = b^2c - bc^2 \]

3. Expand \( ca(c-a) \):
\[ ca(c-a) = c^2a - ca^2 \]

Now, we can combine all these expanded terms:
\[ ab(a-b) + bc(b-c) + ca(c-a) = (a^2b - ab^2) + (b^2c - bc^2) + (c^2a - ca^2) \]

### Step 2: Combine like terms
Next, we combine all the terms:
\[ = a^2b - ab^2 + b^2c - bc^2 + c^2a - ca^2 \]

### Step 3: Rearranging the terms
Now, we rearrange the terms for clarity:
\[ = a^2b + b^2c + c^2a - (ab^2 + bc^2 + ca^2) \]

### Step 4: Factor the expression
To factor the expression, we can look for patterns. We can group the terms:
\[ = a^2b - ab^2 + b^2c - bc^2 + c^2a - ca^2 \]

We can factor by grouping:
\[ = ab(a-b) + bc(b-c) + ca(c-a) \]

### Step 5: Final factorization
The expression can be factored as:
\[ = (a-b)(b-c)(c-a) \]

Thus, the final result is:
\[ = (a-b)(b-c)(c-a) \]