# How to factor intricate polynomial $ab^3 - a^3b + a^3c -ac^3 +bc^3 - b^3c$ [duplicate]

I would like to know how to factor the following polynomial. $$ab^3 - a^3b + a^3c -ac^3 +bc^3 - b^3c$$ What is the method I should use to factor it? If anyone could help.. Thanks in advance.

• Try $b=a$ and... Jul 30, 2015 at 11:21
• Why do you call this a polynomial? Is it a polynomial in the three variables $a,b,c$? To see mathlove's point, think of it as a polynomial in the single variable $a$, with $b,c$ constants. Notice that polynomial has a root $b$. Jul 30, 2015 at 11:47

$ab^3 - a^3b + a^3c -ac^3 +bc^3 - b^3c =$

$ab^3 - a^3b + a^3c -b^3c -ac^3 + bc^3=$

$ab(b^2-a^2)+c(a^3-b^3)-c^3(a-b)=$

$ab(a+b)(b-a)+c(a-b)(a^2+ab+b^2)-c^3(a-b)=$

$(a-b)[-ab(a+b)+c(a^2+ab+b^2)-c^3]=$

$(a-b)(-a^2b-ab^2+ca^2+cab+cb^2-c^3)=$

$(a-b)(ca^2-a^2b+cab-ab^2+cb^2-c^3)=$

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

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

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

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

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

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

The suggestion of replacing $b$ with $a$ was the following: if in such an expression when you replace $b$ with $a$ you get a $0$ then one of the factors may be $(a-b)$.

In the future if you want an idea of the factors that appear you could write on wolframalpha factor$(ab^3 - a^3b + a^3c -ac^3 +bc^3 - b^3c)$

\begin{align*} ab^3 - a^3b + a^3c -ac^3 +bc^3 - b^3c&=-a^3(b-c)+b^3(a-c)+c^3(b-a)\\ &=-a^3(b-c)+b^3(a-c)+c^3[(b-c)+(c-a)]\\ &=(c^3-a^3)(b-c)+(b^3-c^3)(a-c)\\ &=(c-a)(c^2+ac+a^2)(b-c)+(b-c)(b^2+bc+c^2)(a-c)\\ &=(a-c)(b-c)(-c^2-ac-a^2+b^2+bc+c^2)\\ &=(a-c)(b-c)(b^2+bc-ac-a^2)\\ &=(a-c)(b-c)[(b+a)(b-a)+c(b-a)]\\ &=(a-c)(b-c)(b-a)(a+b+c)\\ &=(a-b)(b-c)(c-a)(a+b+c)\\ \end{align*}

• You can continue - $b^2-a^2+bc-ac=(b-a)[a+b+c]$. Jul 30, 2015 at 11:48

Let $$f(a)=ab^3-a^3b+a^3c-ac^3+bc^3-b^3c.$$ If $a=b$, then $f(a)=0$. Therefore $(a-b)\mid f(a)$. Analogously $(a-c)\mid f(a)$; so, $(a-b)(a-c)\mid f(a)$. But we can take it directly: $$ab^3-a^3b+a^3c-ac^3+bc^3-b^3c = ab(b^2-a^2) + c^3(b-a) - c(b^3-a^3) =\\= ab(b-a)(b+a)+c^3(b-a)-c(b-a)(b^2+ba+a^2) =\\= (b-a)[ab(b+a)+c^3-cb^2-cab-ca^2]$$ $$ab(b+a)+c^3-cb^2-cab-ca^2=ab(a+b) + c(c^2-a^2) - cb(a+b) =\\= (a-c)b(a+b) - c(a-c)(a+c)=(a-c)[b(a+b)-c(a+c)]$$ And then $$b(a+b)-c(a+c) = ba+b^2 - ca - c^2 = a(b-c) + (b^2-c^2) =\\= a(b-c)+(b+c)(b-c)=(b-c)(a+b+c).$$ So, $$ab^3-a^3b+a^3c-ac^3+bc^3-b^3c=(b-a)(a-c)(b-c)(a+b+c)$$