1. **State the problem:** Solve the equation $$\sqrt[3]{x+1}+\sqrt[3]{-x+1}=2$$. 2. **Set variables:** Let $$a=\sqrt[3]{x+1}$$ and $$b=\sqrt[3]{-x+1}$$. 3. **Rewrite the equation in terms of \(a\) and \(b\):** $$a+b=2$$ 4. **Express the cubes:** Since $$a^3 = x+1$$ and $$b^3 = -x+1$$, adding these gives: $$a^3 + b^3 = (x+1) + (-x+1) = 2$$ 5. **Use the identity for sum of cubes:** $$a^3 + b^3 = (a+b)^3 - 3ab(a+b)$$ Substitute $$a+b=2$$: $$2 = 2^3 - 3ab \cdot 2 = 8 - 6ab$$ 6. **Solve for $$ab$$:** $$2 = 8 - 6ab \implies 6ab = 8 - 2 = 6 \implies ab = 1$$ 7. **Now, $$a+b=2$$ and $$ab=1$$. This means $$a$$ and $$b$$ are roots of the quadratic equation: $$t^2 - (a+b)t + ab = 0 \implies t^2 - 2t + 1 = 0$$ 8. **Solve the quadratic:** $$t^2 - 2t + 1 = (t-1)^2 = 0$$ Thus, $$a = b = 1$$. 9. **Recall definitions:** $$a = \sqrt[3]{x+1} = 1 \implies x+1 = 1^3 = 1 \implies x = 0$$ 10. **Check the solution for $$b$$:** $$b = \sqrt[3]{-x+1} = \sqrt[3]{-0+1} = \sqrt[3]{1} = 1$$, which confirms consistency. **Answer:** $$\boxed{0}$$