Factorize:

Question:

Factorize:

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

 

Solution:

We have:

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

Put $a(b-c)=x$

$b(c-a)=y$

$c(a-b)=z$

Here,

$x+y+z=a(b-c)+b(c-a)+c(a-b)$

$=a b-a c+b c-a b+a c-b c$

$=0$

Thus, we have :

$a^{3}(b-c)^{3}+b^{3}(c-a)^{3}+c^{3}(a-b)^{3}=x^{3}+y^{3}+z^{3}$

$=3 x y z \quad\left[\right.$ When $\left.x+y+z=0, x^{3}+y^{3}+z^{3}=3 x y z .\right]$

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

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

 

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