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.
