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

The literal $a$ is multiplied by the square of $b$ minus $c$ squared, the literal $b$ is multiplied by the $c$ squared minus square of $a$ and the literal $c$ is multiplied by the $a$ square minus $b$ squared.

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

This algebraic expression should be factored in this problem. So, let’s learn how to factorize (or factorise) the given polynomial.

### Multiply the difference by the factor

There is nothing to separate from the expression. So, it is not possible to factorise the given polynomial.

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

Hence, multiply the difference of the squares in each term by corresponding factor by using the distributive property of multiplication over subtraction.

$=\,\,\,$ $a \times b^2$ $-$ $a \times c^2$ $+$ $b \times c^2$ $-$ $b \times a^2$ $+$ $c \times a^2$ $-$ $c \times b^2$

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

### Write the Terms in a convenient order

The six terms in the algebraic expression should be written in a convenient order on the basis of the common factors.

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

### Factoring the expression by Grouping

There is a common factor in every two terms of the expression. Hence, write every two terms as a group and then implement the factorization of the expression by grouping.

$=\,\,\,$ $(ab^2-ba^2)$ $+$ $(bc^2-ac^2)$ $+$ $(ca^2-cb^2)$

$=\,\,\,$ $(ab \times b-ba \times a)$ $+$ $(b \times c^2-a \times c^2)$ $+$ $(c \times a^2-c \times b^2)$

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

Now, take the factor common from every group in this algebraic expression.

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

The difference of the squares can be expressed in factor form by using the difference rule of the squares.

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

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

$=\,\,\,$ $(a-b) \times \big(-ab-c^2+c \times (a+b)\big)$

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

In this expression, a-b is a factor in each term. So, it can be taken out common from all the terms for factorising the expression.

$=\,\,\,$ $(a-b)(-ab-c^2+ca+cb)$

The terms in the second factor can be written as two groups on the basis of the common factor and it is useful for factorizing the above algebraic expression further.

$=\,\,\,$ $(a-b)\big((cb-ab)+(ca-c^2)\big)$

$=\,\,\,$ $(a-b)\big((c \times b-a \times b)+(c \times a-c \times c)\big)$

Finally, use the taking out the common factor method to finish the factorization of the given polynomial.

$=\,\,\,$ $(a-b)\big(b \times (c-a)+c \times (a-c)\big)$

$=\,\,\,$ $(a-b)\big(b \times (c-a)-c \times (c-a)\big)$

$=\,\,\,$ $(a-b)(c-a) \times (b-c)$

$=\,\,\,$ $(a-b)(b-c) \times (c-a)$

$=\,\,\,$ $(a-b)(b-c)(c-a)$

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