How to factor $(a +b+c) ^5 -(a^5+ b^5 + c^5)$?

Question

How do you factor $(a +b+c) ^5 -(a^5+ b^5 + c^5)$?

The expression in $a,b,c$ is homogeneous.

Answer 1

Hint: Think about $b,c$ as fixed constants. Let $$P(X)=(X +b+c) ^5 -(X^5+ b^5 + c^5)$$

This is a fourth degree polynomial in $X$. It is trivial to see two roots: $$X=-b \\ X=-c$$

Therefore, $$P(X)=(X+b)(X+c) Q(X)$$ for some monic quadratic $Q(X)$. $Q(X)$ can be easily found via long division.

Answer 2

Make use of the cubic polynomial factorization

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

Then,

$$(a+b+c)^5-(a^5+b^5+c^5) = 3(a+b)(b+c)(c+a)(a+b+c)^2 + I\tag 1$$ where $$\begin{array} & I &= (a^3+b^3+c^3)(a+b+c)^2 -(a^5+b^5+c^5) \\ & =a^3[(b+c)^2+2a(b+c)]+ b^3[(c+a)^2+2b(c+a)]+ c^3[(a+b)^2+2c(a+b)]\\ & =a^2(b+c)[2(a+b)(a+c)-(bc+ab+ac)-bc] \\ & \>\>\>\>+b^2(a+c)[2(b+a)(b+c)-(bc+ab+ac)-ac] \\ & \>\>\>\>+c^2(a+b)[2(c+a)(c+b)-(bc+ab+ac)-ab] \\ & =2(a+b)(b+c)(c+a)(a^2+b^2+c^2) \\ & \>\>\>\>- (bc+ab+ac)[a^2(b+c)+b^2(a+c)+c^2(a+b)+2abc] \\ & =(a+b)(b+c)(c+a)[(2a^2+2b^2+2c^2) -(bc+ab+ac)] \\ \end{array}$$

Substitute $I$ into (1) to obtain the factorization, $$(a+b+c)^5-(a^5+b^5+c^5) = 5(a+b)(b+c)(c+a)(a^2+b^2+c^2+ab+bc+ca)$$