# Factorize the expression $a(b^2 -c^2)$ $+$ $b(c^2 -a^2)$ $+$ $c(a^2 -b^2)$

$a(b^2 -c^2) + b(c^2 -a^2) + c(a^2 -b^2)$ is an algebraic expression in terms of literals $a$, $b$ and $c$.

Multiply each binomial expression by its respective multiplying algebraic factor.

$= ab^2 -ac^2 + bc^2 -ba^2 + ca^2 -cb^2$

Arrange the terms in a comfortable order.

$= ab^2 -ba^2 -ac^2 + bc^2 + ca^2 -cb^2$

Take $ab$ common from fist two terms, also take $c^2$ common from next two terms and also take $c$ common from the remaining two algebraic terms in this algebraic expression.

$= ab(b -a) +c^2(-a + b) + c(a^2 -b^2)$

$= -ab(a-b) -c^2(a-b) + c(a^2 -b^2)$

$= -ab(a-b) -c^2(a-b) + c(a+b)(a-b)$

Now, take $a-b$ common from the three terms of this expression.

$= (a-b)[-ab -c^2 + c(a+b)]$

Multiply the literal factor $c$ with the binomial $a+b$.

$= (a-b)[-ab -c^2 + ca + cb]$

$= (a-b)[-ab + cb -c^2 + ca]$

Take the literal $b$ common from first two terms and also take $c$ common from remaining two terms.

$= (a-b)[b(-a+c) + c(-c+a)]$

$= (a-b)[b(c-a) -c(c-a)]$

Take $c-a$ common from both terms.

$= (a-b)[(c-a)(b-c)]$

$\therefore \,\,\, a(b^2 -c^2) + b(c^2 -a^2) + c(a^2 -b^2) = (a-b)(b-c)(c-a)$

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