Question

The expression \((a-b)^3 + (b-c)^3 + (c-a)^3\) can be factorized as:

• A. \((a - b)(b - c)(c - a)\)
• B. \(3(a - b)(b - c)(c - a)\)
• C. \(3(a + b)(b - c)(c + a)\)
• D. None of the option is correct

Answer 1

The Correct Option is B

To factorize \((a-b)^3 + (b-c)^3 + (c-a)^3\), we use the identity for the sum of cubes in a cyclic form.
The identity is:\[x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)\]However, in our case, we have \((a-b)^3 + (b-c)^3 + (c-a)^3\) and not a sum of three cubes directly with \(x + y + z = 0\). Therefore, we can apply a known algebraic identity:
\[(a-b)^3 + (b-c)^3 + (c-a)^3 = 3(a - b)(b - c)(c - a)\]
This identity works when \(a + b + c = 0\) (which is a special case for our cyclic form where each term cancels the others).
Therefore, the factorization of the given expression is:
\[(a-b)^3 + (b-c)^3 + (c-a)^3 = 3(a - b)(b - c)(c - a)\]
Thus, the correct answer is:
Option B. \(3(a - b)(b - c)(c - a)\)