(a+b+c)(a+b-c)(b+c-a)(c+a-b)

3 min read Jun 16, 2024

Factoring the Expression (a+b+c)(a+b-c)(b+c-a)(c+a-b)

This expression appears complex at first glance, but it can be simplified significantly using a clever factorization technique. Let's explore how to break it down:

Understanding the Pattern

Notice that the terms within the parentheses follow a specific pattern:

First factor: (a + b + c)
Second factor: (a + b - c)
Third factor: (b + c - a)
Fourth factor: (c + a - b)

Each factor has two positive terms and one negative term, and the negative term changes its sign systematically. This pattern is crucial for the factorization.

Applying the Difference of Squares

Let's group the factors strategically:

Group 1: (a + b + c)(a + b - c)
Group 2: (b + c - a)(c + a - b)

Now, we can apply the difference of squares pattern:

(x + y)(x - y) = x² - y²

Applying this to each group:

Group 1: (a + b)² - c²
Group 2: (b + c)² - a²

Expanding and Simplifying

Let's expand the squares and rearrange the terms:

Group 1: a² + 2ab + b² - c²
Group 2: b² + 2bc + c² - a²

Now, we have:

**(a² + 2ab + b² - c²)(b² + 2bc + c² - a²) **

Observe that the terms involving a² and b² cancel out:

(2ab + 2bc + 2ac)(2ab + 2bc + 2ac)

Finally, we can factor out a common factor of 2:

(2)(ab + bc + ac)(2)(ab + bc + ac)

Therefore, the simplified expression is: 4(ab + bc + ac)²

Conclusion

By recognizing the pattern in the expression and applying the difference of squares, we successfully factored the expression into a much simpler form: 4(ab + bc + ac)². This method highlights the importance of identifying patterns and applying algebraic identities to simplify complex expressions.