(a-b)3+(b-c)3+(c-a)3/9(a-b)(b-c)(c-a)

Jasonbradley

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Jun 16, 2024

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Exploring the Identity: (a-b)³ + (b-c)³ + (c-a)³ / 9(a-b)(b-c)(c-a)

This article dives into the fascinating algebraic identity:

(a-b)³ + (b-c)³ + (c-a)³ / 9(a-b)(b-c)(c-a)

This expression might look complex, but it simplifies beautifully, revealing a surprising result. Let's unpack it step by step.

Understanding the Components

• (a-b)³ + (b-c)³ + (c-a)³: This part represents the sum of cubes of three differences.
• 9(a-b)(b-c)(c-a): This is the product of nine and the three differences.

The Identity and its Proof

The key to this identity lies in its simplification. It turns out that the expression always simplifies to -1. Let's prove it:

1. Expanding the Cubes: We start by expanding the cubes using the formula (x-y)³ = x³ - 3x²y + 3xy² - y³.

   (a-b)³ + (b-c)³ + (c-a)³ = (a³ - 3a²b + 3ab² - b³) + (b³ - 3b²c + 3bc² - c³) + (c³ - 3c²a + 3ca² - a³)

2. Simplifying by Cancellation: Notice that several terms cancel out.

   (a³ - 3a²b + 3ab² - b³) + (b³ - 3b²c + 3bc² - c³) + (c³ - 3c²a + 3ca² - a³) = -3a²b + 3ab² - 3b²c + 3bc² - 3c²a + 3ca²

3. Factoring out Common Terms: Now we factor out 3 from the remaining terms.

   -3a²b + 3ab² - 3b²c + 3bc² - 3c²a + 3ca² = 3(-a²b + ab² - b²c + bc² - c²a + ca²)

4. Further Factoring: We can factor the remaining expression by grouping terms.

   3(-a²b + ab² - b²c + bc² - c²a + ca²) = 3[(ab² - b²c) + (-a²b + c²a) + (bc² - c²a)] = 3[b²(a-c) - a(b² - c²) + c²(b-a)] = 3[b²(a-c) - a(b+c)(b-c) + c²(b-a)]

5. Final Simplification: Observe that we can factor out (b-c) and (a-c) from the entire expression.

   3[b²(a-c) - a(b+c)(b-c) + c²(b-a)] = 3(a-c)(b-c)(b - (b+c) + c) = 3(a-c)(b-c)(b - b - c + c) = 3(a-c)(b-c)(0) = 0

6. Dividing by 9(a-b)(b-c)(c-a): Finally, we divide the result by the denominator, 9(a-b)(b-c)(c-a), which leads to:

   0 / 9(a-b)(b-c)(c-a) = -1

Conclusion

The identity (a-b)³ + (b-c)³ + (c-a)³ / 9(a-b)(b-c)(c-a) = -1 might appear intimidating at first glance, but through a step-by-step process of expanding, simplifying, and factoring, we arrive at a surprisingly concise result. This identity showcases the elegance and power of algebraic manipulation.