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The expression vanishes when $$a = -b$$, since it becomes $$c^3 - (-b^3 + b^3 + c^3) = 0$$, so $$(a + b)$$ is a factor, and by symmetry so are $$(b + c)$$ and $$(c + a)$$. Being of degree 3, it equals $$k(a + b)(b + c)(c + a)$$. Putting $$a = b = c = 1$$ gives $$27 - 3 = 24 = 8k$$, so $$k = 3$$ and the factorization is $$3(a + b)(b + c)(c + a)$$.
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