The expression [tex](a-b)^3+(b-c)^3+(c-a)^3[/tex] can be factorized as:

A. [tex](a-b)(b-c)(c-a)[/tex]

B. [tex]3(a-b)(b-c)(c-a)[/tex]

C. [tex]-3(a-b)(b-c)[/tex]

D. [tex](a+b+c)(c - a)\left(a^2+b^2+c^2-ab-bc-ca\right)[/tex]



Answer :

To factorize the expression [tex]\((a-b)^3+(b-c)^3+(c-a)^3\)[/tex], let's follow a detailed, step-by-step approach:

1. Understanding the Expression:
We are given the expression [tex]\((a-b)^3 + (b-c)^3 + (c-a)^3\)[/tex] and asked to factorize it.

2. Sum of Cubes Identity:
There's an identity in algebra known as the sum of cubes identity:
[tex]\[ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) \][/tex]
In this identity, if [tex]\(x + y + z = 0\)[/tex], then the expression simplifies to:
[tex]\[ x^3 + y^3 + z^3 = 3xyz \][/tex]

3. Assigning Values:
In our expression, let:
[tex]\[ x = (a - b), \quad y = (b - c), \quad z = (c - a) \][/tex]

4. Sum of x, y, and z:
Add the terms [tex]\(x, y, z\)[/tex] to check if their sum equals zero:
[tex]\[ (a - b) + (b - c) + (c - a) \][/tex]
This simplifies to:
[tex]\[ a - b + b - c + c - a = 0 \][/tex]
Since [tex]\(x + y + z = 0\)[/tex], we can use the simplification mentioned earlier.

5. Simplified Sum of Cubes:
According to the identity, when [tex]\(x + y + z = 0\)[/tex],
[tex]\[ x^3 + y^3 + z^3 = 3xyz \][/tex]
Thus, substituting back:
[tex]\[ (a - b)^3 + (b - c)^3 + (c - a)^3 = 3(a - b)(b - c)(c - a) \][/tex]

Therefore, the factorized form of the given expression is:

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

This corresponds to:

(B) [tex]\(\boxed{3(a-b)(b-c)(c-a)}\)[/tex]