1. **Problem statement:** Given a triangle with sides $a$, $b$, and $c$, and the relation $$\frac{1}{a+b} + \frac{1}{b+c} = \frac{3}{a+b+c},$$ find the angle opposite side $b$. 2. **Formula and rules:** In any triangle, the Law of Cosines relates the sides and the angle opposite a side: $$b^2 = a^2 + c^2 - 2ac \cos B,$$ where $B$ is the angle opposite side $b$. 3. **Rewrite the given relation:** Multiply both sides by $(a+b)(b+c)(a+b+c)$ to clear denominators: $$(b+c)(a+b+c) + (a+b)(a+b+c) = 3(a+b)(b+c).$$ 4. **Expand terms:** $$(b+c)(a+b+c) = (b+c)(a+b+c) = (b+c)(a+b+c),$$ which expands to $$ (b+c)(a+b+c) = (b+c)(a+b+c) = (b+c)(a+b+c).$$ Actually, let's expand carefully: $$(b+c)(a+b+c) = (b+c)a + (b+c)b + (b+c)c = a(b+c) + b(b+c) + c(b+c) = ab + ac + b^2 + bc + bc + c^2 = ab + ac + b^2 + 2bc + c^2.$$ Similarly, $$(a+b)(a+b+c) = a(a+b+c) + b(a+b+c) = a^2 + ab + ac + ab + b^2 + bc = a^2 + 2ab + ac + b^2 + bc.$$ 5. **Sum the two expansions:** $$ab + ac + b^2 + 2bc + c^2 + a^2 + 2ab + ac + b^2 + bc = 3(a+b)(b+c).$$ Simplify left side: $$a^2 + 3ab + 2ac + 2b^2 + 3bc + c^2 = 3(a+b)(b+c).$$ 6. **Expand right side:** $$3(a+b)(b+c) = 3(ab + ac + b^2 + bc) = 3ab + 3ac + 3b^2 + 3bc.$$ 7. **Set equation:** $$a^2 + 3ab + 2ac + 2b^2 + 3bc + c^2 = 3ab + 3ac + 3b^2 + 3bc.$$ 8. **Bring all terms to one side:** $$a^2 + 3ab + 2ac + 2b^2 + 3bc + c^2 - 3ab - 3ac - 3b^2 - 3bc = 0,$$ which simplifies to $$a^2 + (3ab - 3ab) + (2ac - 3ac) + (2b^2 - 3b^2) + (3bc - 3bc) + c^2 = 0,$$ $$a^2 - ac - b^2 + c^2 = 0.$$ 9. **Rewrite:** $$a^2 + c^2 - ac - b^2 = 0,$$ or $$a^2 + c^2 - b^2 = ac.$$ 10. **Recall Law of Cosines:** $$b^2 = a^2 + c^2 - 2ac \cos B,$$ so $$a^2 + c^2 - b^2 = 2ac \cos B.$$ 11. **Substitute into previous equation:** $$2ac \cos B = ac,$$ which gives $$\cos B = \frac{ac}{2ac} = \frac{1}{2}.$$ 12. **Find angle $B$:** $$B = \arccos \frac{1}{2} = 60^\circ.$$ **Final answer:** The angle opposite side $b$ is $60^\circ$.