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

Answer 2

Since the claimed factorisation should hold for any values of \(a\), \(b\), \(c\), we can test it by substituting simple numbers rather than deriving the identity algebraically.

Let \(a=1\), \(b=2\), \(c=3\). Then \(a-b=-1\), \(b-c=-1\), \(c-a=2\), so the left side is:
\[(-1)^{3}+(-1)^{3}+2^{3} = -1-1+8 = 6\]

1. Option A \((a-b)(b-c)(c-a)\): \((-1)(-1)(2)=2\), which does not equal 6.
2. Option B \(3(a-b)(b-c)(c-a)\): \(3(-1)(-1)(2)=6\), which matches the left side exactly.
3. Option C \(3(a+b)(b-c)(c+a)\): \(3(3)(-1)(4)=-36\), which does not equal 6.
4. Option D (None of the option is correct): Ruled out since Option B matches exactly.

A second check with \(a=2\), \(b=0\), \(c=1\) gives the left side as \(8-1-1=6\), and Option B again gives \(3(2)(-1)(-1)=6\), confirming the identity.

Hence, the correct answer is Option B: \(3(a-b)(b-c)(c-a)\).