`a ^(2) - b ^(2) - c ^(2) + 2 bc + a + b - c` when factorised equals

Options

A. `(a - b - c) (a - b + c + 1)`
B. `(a + b -c ) (a - b + c + 1)`
C. `(a - b + c) ( a - b + c + 1)`
D. `(a + b + c) (a - b + c + 1)`

Answer

To factor the expression \( a^2 - b^2 - c^2 + 2bc + a + b - c \), we can follow these steps:

### Step 1: Rearrange the expression
We start by rearranging the terms for clarity:
\[ a^2 + a - b^2 - c^2 + 2bc + b - c \]

### Step 2: Group the terms
Next, we can group the terms that can be factored together:
\[ (a^2 + a) + (-b^2 + 2bc - c^2) + (b - c) \]

### Step 3: Factor the first group
From the first group \( a^2 + a \), we can factor out \( a \):
\[ a(a + 1) + (-b^2 + 2bc - c^2) + (b - c) \]

### Step 4: Recognize a perfect square
The second group \( -b^2 + 2bc - c^2 \) can be rewritten as:
\[ -(b^2 - 2bc + c^2) = -(b - c)^2 \]
So now we have:
\[ a(a + 1) - (b - c)^2 + (b - c) \]

### Step 5: Combine the last two groups
Now we can combine the last two groups:
\[ a(a + 1) + (b - c) - (b - c)^2 \]

### Step 6: Factor the entire expression
We can recognize that we can factor out \( (b - c) \) from the last two terms:
\[ = a(a + 1) + (b - c)(1 - (b - c)) \]
This simplifies to:
\[ = a(a + 1) + (b - c)(1 - b + c) \]

### Step 7: Final factorization
Now we can factor the entire expression:
\[ = (a + b - c)(a - b + c + 1) \]

Thus, the factorized form of the expression \( a^2 - b^2 - c^2 + 2bc + a + b - c \) is:
\[ (a + b - c)(a - b + c + 1) \]