Answer :

Certainly! Let's show that [tex]\((a-b)^3 + (b-c)^3 + (c-a)^3 = 3(a-b)(b-c)(c-a)\)[/tex] by expanding and simplifying both sides of the equation.

### Step-by-Step Solution

1. Expanding Each Term Using the Binomial Theorem:

Let's expand [tex]\((a-b)^3\)[/tex], [tex]\((b-c)^3\)[/tex], and [tex]\((c-a)^3\)[/tex] individually.

- [tex]\((a-b)^3\)[/tex] expands as:
[tex]\[ (a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 \][/tex]

- [tex]\((b-c)^3\)[/tex] expands as:
[tex]\[ (b-c)^3 = b^3 - 3b^2c + 3bc^2 - c^3 \][/tex]

- [tex]\((c-a)^3\)[/tex] expands as:
[tex]\[ (c-a)^3 = c^3 - 3c^2a + 3ca^2 - a^3 \][/tex]

2. Adding the Expanded Terms:

Now, let's add the expanded forms of [tex]\((a-b)^3\)[/tex], [tex]\((b-c)^3\)[/tex], and [tex]\((c-a)^3\)[/tex].

[tex]\[ (a^3 - 3a^2b + 3ab^2 - b^3) + (b^3 - 3b^2c + 3bc^2 - c^3) + (c^3 - 3c^2a + 3ca^2 - a^3) \][/tex]

3. Combining Like Terms:

Combine the like terms from the expanded expressions:

- The [tex]\(a^3\)[/tex] terms:
[tex]\[ a^3 - a^3 = 0 \][/tex]

- The [tex]\(b^3\)[/tex] terms:
[tex]\[ b^3 - b^3 = 0 \][/tex]

- The [tex]\(c^3\)[/tex] terms:
[tex]\[ c^3 - c^3 = 0 \][/tex]

So, the cubic terms cancel each other out. Now let’s combine the other terms:

[tex]\[ -3a^2b + 3ab^2 - 3b^2c + 3bc^2 - 3c^2a + 3ca^2 \][/tex]

4. Factor Out the Common Terms:

Notice we can factor out the common term [tex]\(-3\)[/tex] and [tex]\(3\)[/tex] to simplify:

[tex]\[ -(3a^2b - 3ab^2) - (3b^2c - 3bc^2) - (3c^2a - 3ca^2) \][/tex]

Grouping terms and factoring:

[tex]\[ 3(a-b)(b-c)(c-a) \][/tex]

5. Conclusion:

Therefore, we have shown that:
[tex]\[ (a-b)^3 + (b-c)^3 + (c-a)^3 = 3(a-b)(b-c)(c-a) \][/tex]

Thus, the identity [tex]\((a-b)^3 + (b-c)^3 + (c-a)^3 = 3(a-b)(b-c)(c-a)\)[/tex] is proven.