Factorize $$a^5(b-c) + b^5(c-a) + c^5(a-b)$$

Call the expression $$E$$. It vanishes when $$a = b$$, $$b = c$$ or $$c = a$$, so it is divisible by $$(a-b)(b-c)(c-a)$$, and since both are alternating the quotient is symmetric of degree 3, say $$-\left[u\sum a^3 + v\sum a^2b + w\,abc\right]$$. Putting $$a = 0$$ gives $$E = bc(b-c)(b+c)(b^2+c^2)$$ and $$(a-b)(b-c)(c-a) = -bc(b-c)$$, so the quotient is $$-(b^3 + b^2c + bc^2 + c^3)$$, giving $$u = v = 1$$. Putting $$a = 1, b = 2, c = 3$$ gives $$E = -180$$ and $$(a-b)(b-c)(c-a) = 2$$, so $$36 + 48 + 6w = 90$$ and $$w = 1$$. Hence $$E = -(a-b)(b-c)(c-a)\left(a^3 + b^3 + c^3 + a^2b + ab^2 + b^2c + bc^2 + c^2a + ca^2 + abc\right)$$.