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]
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]
