a) Use the remainder theorem to prove that $(a+b+c)$ is a factor of $(a^3+b^3+c^3-3abc)$ . Then find the other factor.

b) Hence factor $(ab^3-ac^3+bc^3-ba^3+ca^3-cb^3)$

So far I have managed to find the other remainder being $(a^2+b^2+c^2-ab-ac-bc)$ but I don't understand how to use this to hence factor the expression.

Also I was reading a solution which factorised the expression without using part a) but how does the second last line turn into the last line?

\begin{align}&\color{white}=(bc^3−cb^3)−(ac^3−ca^3)+(ab^3−ba^3)\\ &=bc(c^2−b^2)−ac(c^2−a^2)+ab(b^2−a^2)\\ &=(c−b)(c+b)bc−ac(c−a)(c+a)+ab(b−a)(b+a)\\ &=−(a−c)(b−c)(a−b)(a+b+c)\end{align}

Thanks in advance

  • $\begingroup$ The second turns into the third by the factorization of each respective difference of squares. $\endgroup$
    – LuuBluum
    Apr 19, 2017 at 4:36
  • 1
    $\begingroup$ Related: 1, 2, 3, 4, 5, 6. $\endgroup$
    – Frenzy Li
    Apr 19, 2017 at 5:38
  • $\begingroup$ I'm just curious, where is this problem taken from? $\endgroup$
    – user49640
    Apr 19, 2017 at 6:44
  • $\begingroup$ @amWhy This is not a duplicate. The other question doesn't ask how to do this in a way that uses the factorization of $a^3 + b^3 + c^3 - 3abc$. $\endgroup$
    – user49640
    Apr 19, 2017 at 19:29

2 Answers 2


The polynomial $$f(a,b,c)=ab^3-ac^3+bc^3-ba^3+ca^3-cb^3$$ is an alternating function of $a$, $b$, $c$. That is it changes sign whenever you swap two variables: $$f(a,b,c)=-f(b,a,c)=-f(c,b,a)=-f(a,c,b).$$ This means that $a-b$, $a-c$ and $b-c$ are all factors (if a polynomial changes sign when you swap $a$ and $b$ then $a-b$ is a factor). Therefore $$f(a,b,c)=(a-b)(a-c)(b-c)g(a,b,c)$$ for some polynomial $g(a,b,c)$. As $f$ is alternating, $g$ is symmetric: $$g(a,b,c)=g(b,a,c)=g(c,b,a)=g(a,c,b).$$ It's clear that $g$ is homogeneous of $a$, $b$ and $c$ of degree one. So $g(a,b,c)=k(a+b+c)$ for some constant $k$. One only needs to compute one coefficient to see that $k=-1$. Therefore $$f(a,b,c)=-(a-b)(a-c)(b-c)(a+b+c).$$

As a follow-up you might try to factorise $$ab^4-ac^4+bc^4-ba^4+ca^4-cb^4$$ or $$a^2b^3-a^2c^3+b^2c^3-b^2a^3+c^2a^3-c^2b^3.$$


Hint (without using the first part): considering it as a $3^{rd}$ degree polynomial in $a\,$:

$$ab^3-ac^3+bc^3-ba^3+ca^3-cb^3 = (c-b)\,a^3+(b^3-c^3)\,a+bc\,(b^2-c^2)$$

It can be easily verified that it has $a=b$ and $a=c$ as roots. Since the term in $a^2$ is missing, the sum of all three roots is $0$ by Vieta's relations, so the third root is $a=-(b+c)\,$. Therefore the polynomial is divisible by $(a-b)(a-c)(a+b+c)\,$. Comparing the leading coefficients, it follows that the scalar multiplier is $c-b\,$, so in the end:

$$ab^3-ac^3+bc^3-ba^3+ca^3-cb^3 = (c-b)(a-b)(a-c)(a+b+c)$$


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