3%2B(b-c)3%2B(c-a)3-is-equal-to.jpeg "(a-b)3+(b-c)3+(c-a)3 Is Equal To")
Exploring the Expansion: (a-b)³ + (b-c)³ + (c-a)³
This article delves into the intriguing algebraic expression (a-b)³ + (b-c)³ + (c-a)³, exploring its simplification and revealing its surprising result.
Expanding the Expression
The first step is to expand the cubes using the algebraic identity: (x + y)³ = x³ + 3x²y + 3xy² + y³.
Applying this to our expression, we get:
(a-b)³ = a³ - 3a²b + 3ab² - b³
(b-c)³ = b³ - 3b²c + 3bc² - c³
(c-a)³ = c³ - 3c²a + 3ca² - a³
Combining Terms
Now, let's combine the terms from each expansion:
(a³ - 3a²b + 3ab² - b³) + (b³ - 3b²c + 3bc² - c³) + (c³ - 3c²a + 3ca² - a³)
Notice that the cubic terms (a³, b³, c³) cancel out. We are left with:
-3a²b + 3ab² - 3b²c + 3bc² - 3c²a + 3ca²
Factoring and the Final Result
The remaining terms can be factored by grouping:
3ab(a - b) - 3bc(b - c) - 3ca(c - a)
Finally, we can factor out a 3 and simplify further:
3 [ab(a - b) - bc(b - c) - ca(c - a)]
Therefore, (a-b)³ + (b-c)³ + (c-a)³ = 3 [ab(a - b) - bc(b - c) - ca(c - a)].
Conclusion
The expansion of (a-b)³ + (b-c)³ + (c-a)³ leads to a simplified expression 3 [ab(a - b) - bc(b - c) - ca(c - a)]. This demonstrates a fascinating algebraic pattern and highlights the power of factorization in simplifying complex expressions.