Factorization of (a − b)³ + (b − c)³ + (c − a)³
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) \((a+b+c)(a^2+b^2+c^2-ab-bc-ca)\)
Solution
Let
\[ x=a-b,\quad y=b-c,\quad z=c-a \]
Then,
\[ x+y+z=(a-b)+(b-c)+(c-a)=0 \]
Using identity:
\[ x^3+y^3+z^3=3xyz \]
\[ (a-b)^3+(b-c)^3+(c-a)^3 \]
\[ =3(a-b)(b-c)(c-a) \]
Therefore,
\[ \boxed{(b)\ 3(a-b)(b-c)(c-a)} \]