(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3/(a-b)^3+(b-c)^3+(c-a)^3

Jasonbradley

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Jun 16, 2024

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Simplifying the Expression (a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3/(a-b)^3+(b-c)^3+(c-a)^3

This expression might look intimidating at first, but with a bit of algebraic manipulation, we can simplify it significantly. Here's how:

Key Observation: The Sum of Cubes

The expression has a form that suggests using the sum of cubes factorization:

• a^3 + b^3 = (a + b)(a^2 - ab + b^2)

Let's apply this to our problem.

1. Factor the Numerator

Notice that the numerator is the sum of cubes:

• (a^2 - b^2)^3 + (b^2 - c^2)^3 + (c^2 - a^2)^3

Let's make a substitution to simplify things. Let:

• x = a^2 - b^2
• y = b^2 - c^2
• z = c^2 - a^2

Now our numerator becomes:

• x^3 + y^3 + z^3

Using the sum of cubes factorization, we get:

• (x + y + z)(x^2 - xy + y^2 + xz - yz + z^2)

2. Factor the Denominator

Similarly, we can factor the denominator using the sum of cubes factorization:

• (a - b)^3 + (b - c)^3 + (c - a)^3

Making substitutions:

• p = a - b
• q = b - c
• r = c - a

We get:

• p^3 + q^3 + r^3 = (p + q + r)(p^2 - pq + q^2 + pr - qr + r^2)

3. Simplify the Expression

Now we can rewrite the original expression:

[(x + y + z)(x^2 - xy + y^2 + xz - yz + z^2)] / [(p + q + r)(p^2 - pq + q^2 + pr - qr + r^2)]

Notice that (x + y + z) = [(a^2 - b^2) + (b^2 - c^2) + (c^2 - a^2)] = 0

And (p + q + r) = [(a - b) + (b - c) + (c - a)] = 0

Since the numerator and denominator are both multiplied by zero, the entire expression simplifies to 0.

Conclusion

The seemingly complex expression (a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3/(a-b)^3+(b-c)^3+(c-a)^3 reduces to 0. This simplification highlights the power of algebraic manipulation and factorization.