Factorize:

Question:

Factorize:

$(a-b)^{3}+(b-c)^{3}+(c-a)^{3}$

Solution:

$(a-b)^{3}+(b-c)^{3}+(c-a)^{3}$

Putting $(a-b)=x,(b-c)=y$ and $(c-a)=z$, we get:

$(a-b)^{3}+(b-c)^{3}+(c-a)^{3}$

$=x^{3}+y^{3}+z^{3} \quad[$ Where $(x+y+z)=(a-b)+(b-c)+(c-a)=0]$

$=3 x y z \quad\left[(x+y+z)=0 \Rightarrow x^{3}+y^{3}+z^{3}=3 x y z\right]$

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