Factoring the Expression (a²-b²)³ + (b²-c²)³ + (c²-a²)³
This article explores the factorization of the expression (a²-b²)³ + (b²-c²)³ + (c²-a²)³. We will employ a combination of algebraic manipulation and the application of a helpful algebraic identity.
Understanding the Problem
The given expression consists of three cubes, each involving the difference of squares. Our goal is to find a factored form for this expression, ideally in terms of simpler factors.
Applying the Sum of Cubes Identity
We can utilize the following identity:
x³ + y³ = (x + y)(x² - xy + y²)
To apply this identity, we need to identify 'x' and 'y' in our expression. Let's set:
- x = (a² - b²)
- y = (b² - c²)
Substituting these values into the identity, we get:
(a² - b²)³ + (b² - c²)³ = [(a² - b²) + (b² - c²)][(a² - b²)² - (a² - b²)(b² - c²) + (b² - c²)²]
Simplifying the first factor:
(a² - b²) + (b² - c²) = a² - c²
Now, let's focus on the second factor. Expanding the squares and simplifying:
(a² - b²)² - (a² - b²)(b² - c²) + (b² - c²)² = (a⁴ - 2a²b² + b⁴) - (a²b² - a²c² - b⁴ + b²c²) + (b⁴ - 2b²c² + c⁴)
Combining like terms:
(a⁴ - 2a²b² + b⁴) - (a²b² - a²c² - b⁴ + b²c²) + (b⁴ - 2b²c² + c⁴) = a⁴ + a²c² + 2b⁴ - 3a²b² - 3b²c² + c⁴
Now, we have:
(a² - b²)³ + (b² - c²)³ = (a² - c²)(a⁴ + a²c² + 2b⁴ - 3a²b² - 3b²c² + c⁴)
Dealing with the Remaining Cube
We still need to incorporate the third cube, (c² - a²)³. Notice that we can rewrite it as:
(c² - a²)³ = - (a² - c²)³
Now, we can substitute this back into our original expression:
(a² - b²)³ + (b² - c²)³ + (c² - a²)³ = (a² - c²)(a⁴ + a²c² + 2b⁴ - 3a²b² - 3b²c² + c⁴) - (a² - c²)³
We can factor out (a² - c²):
(a² - c²) [(a⁴ + a²c² + 2b⁴ - 3a²b² - 3b²c² + c⁴) - (a² - c²)²]
Finally, expanding and simplifying the expression inside the brackets:
(a² - c²) [(a⁴ + a²c² + 2b⁴ - 3a²b² - 3b²c² + c⁴) - (a⁴ - 2a²c² + c⁴)]
(a² - c²) [3a²c² + 2b⁴ - 3a²b² - 3b²c²]
Therefore, the completely factored form of the expression is:
**(a² - b²)³ + (b² - c²)³ + (c² - a²)³ = (a² - c²)(3a²c² + 2b⁴ - 3a²b² - 3b²c²) **
Conclusion
By applying the sum of cubes identity and carefully manipulating algebraic expressions, we successfully factored the given expression. The final result is a product of two factors, revealing the underlying structure of the original expression.