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. $3(a - b)(b - c)(c - a)$