Factorise :
` b^(2) + c^(2) + 2bc - a^(2)`

Answer

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

### Step 1: Recognize a Perfect Square
We start with the expression:
\[ b^2 + c^2 + 2bc - a^2 \]
Notice that \( b^2 + c^2 + 2bc \) can be rewritten using the perfect square formula:
\[ b^2 + c^2 + 2bc = (b + c)^2 \]
Thus, we can rewrite the original expression as:
\[ (b + c)^2 - a^2 \]

### Step 2: Apply the Difference of Squares Formula
Now we have a difference of squares, which can be factored using the formula \( x^2 - y^2 = (x - y)(x + y) \). Here, let \( x = (b + c) \) and \( y = a \). Therefore, we can write:
\[ (b + c)^2 - a^2 = ((b + c) - a)((b + c) + a) \]

### Step 3: Simplify the Factors
Now we simplify the factors:
1. The first factor is:
\[ (b + c) - a = b + c - a \]
2. The second factor is:
\[ (b + c) + a = b + c + a \]

### Final Factorized Form
Putting it all together, we have:
\[ b^2 + c^2 + 2bc - a^2 = (b + c - a)(b + c + a) \]

### Summary of the Factorization
Thus, the factorization of the expression \( b^2 + c^2 + 2bc - a^2 \) is:
\[ (b + c - a)(b + c + a) \]

---