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.

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.

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

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

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

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

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.

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

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

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

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

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

Latest Math Topics

Dec 13, 2023

Jul 20, 2023

Jun 26, 2023

Latest Math Problems

Jan 30, 2024

Oct 15, 2023

Copyright © 2012 - 2023 Math Doubts, All Rights Reserved