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Quantitative Ability and Data Interpretation Question on Algebra

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

A

(a−b)(b−c)(c−a)(a - b)(b - c)(c - a)

B

3(a−b)(b−c)(c−a)3(a - b)(b - c)(c - a)

C

3(a+b)(b−c)(c+a)3(a + b)(b - c)(c + a)

D

None of the option is correct

Answer

3(a−b)(b−c)(c−a)3(a - b)(b - c)(c - a)

Explanation

Solution

To factorize (a−b)3+(b−c)3+(c−a)3(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:x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)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(a-b)^3 + (b-c)^3 + (c-a)^3 and not a sum of three cubes directly with x+y+z=0x + 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)(a-b)^3 + (b-c)^3 + (c-a)^3 = 3(a - b)(b - c)(c - a)
This identity works when a+b+c=0a + 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)(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)3(a - b)(b - c)(c - a)