(x-a)^3+(x-b)^3+(x-c)^3-3(x-a)(x-b)(x-c)

Jasonbradley

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Jun 17, 2024

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Factoring the Expression (x-a)³ + (x-b)³ + (x-c)³ - 3(x-a)(x-b)(x-c)

This expression appears frequently in algebra and is a special case of a more general factorization pattern. Let's explore how to factor this expression:

The Key Identity

The key to factoring this expression lies in the following algebraic identity:

a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - ac - bc)

This identity can be proved using various algebraic manipulations.

Applying the Identity

Now, let's apply this identity to our expression:

1. Identify a, b, and c: In our expression, a = (x-a), b = (x-b), and c = (x-c).

2. Substitute: Substitute these values into the identity:

   (x-a)³ + (x-b)³ + (x-c)³ - 3(x-a)(x-b)(x-c) = [(x-a) + (x-b) + (x-c)][(x-a)² + (x-b)² + (x-c)² - (x-a)(x-b) - (x-a)(x-c) - (x-b)(x-c)]

3. Simplify: Simplify the expression:

   = (3x - a - b - c) [(x² - 2ax + a²) + (x² - 2bx + b²) + (x² - 2cx + c²) - (x² - ax - bx + ab) - (x² - ax - cx + ac) - (x² - bx - cx + bc)]

   = (3x - a - b - c)(3x² - 3ax - 3bx - 3cx + a² + b² + c² - ab - ac - bc)

   = (3x - a - b - c)(x² - ax - bx - cx + a² + b² + c² - ab - ac - bc)

Conclusion

Therefore, the factored form of (x-a)³ + (x-b)³ + (x-c)³ - 3(x-a)(x-b)(x-c) is (3x - a - b - c)(x² - ax - bx - cx + a² + b² + c² - ab - ac - bc).

This factorization can be useful in various mathematical problems, especially when dealing with cubic equations and simplifying expressions.