Help with factoring $a^4(b-c)+b^4(c-a)+c^4(a-b)$

Question

I want to factor $a^4(b-c)+b^4(c-a)+c^4(a-b)$. I found a similar question on the site, but still don't know how to factor it. from Macavity's answer I realized it has the factor of $(a-b)(a-c)(b-c)$. as mentioned in the answer I should find a quadratic that after multiplying on this factor getting the initial expression. I expanded these expressions:

$$a^4b-a^4c-b^4c-b^4a+c^4a-c^4b=(a^2b-a^2c-b^2a+b^2c+ac^2-bc^2)\times\underbrace{(\quad\quad\quad)}_\text{quadratic expression}$$ I don't know how to find it.

Answer 1

See both polynomials as polynomials in $a$; the numerator is$$(b-c)a^4+(-b^4+c^4)a+b^4c-bc^4\tag1$$and the denominator is$$(b-c)a^2+(-b^2+c^2)a+b^2c-bc^2.\tag2$$Now, apply long polynomial division to $(1)$ and $(2)$, and you will get$$a^2+(b+c)a+b^2+bc+c^2.$$

Answer 2

Let $f(a,b,c)$ be the polynomial to factor. Write $f(a,b,c)= g(a,b,c)(a-b)(a-c)(b-c)$ with $g(a,b,c)=r_1a^2+r_2ab+r_3ac+r_4b^2+r_5bc+r_6c^2$. Compare coefficients to obtain $$ g(a,b,c)=a^2 + ab + ac + b^2 + bc + c^2. $$ One can make this more efficient by arguing that the coefficients for $ac,ab,bc$ must be equal and the ones for $a^2,b^2,c^2$ as well.

So we have the factorization $$ f(a,b,c)=(a^2 + ab + ac + b^2 + bc + c^2)(a - b)(a - c)(b - c) $$

Answer 3

I think the following method is better :we have $$a^4(b-a+a-c)+b^4(c-a)+c^4(a-b)$$ $$=(a^4-c^4)(b-a)+(a-c)(a^4-b^4)$$ $$=(a-c)(b-a)(a^3+ac^2+a^2c+c^3)+(a-c)(a-b)(a^3+ab^2+a^2b+b^3)$$ $$=(a-c)(b-a)(\color{blue}{ac^2-ab^2}+\color{red}{a^2c-a^2b}+\color{green}{c^3-b^3})$$ can you end it now?

Note we just used well known identities $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+...b^{n-1})$