1. **State the problem:** Rationalize the denominator of the fraction $$\frac{1}{\sqrt[3]{5} + \sqrt[3]{7} + 2}$$ where the denominator has two cube roots and one rational part. 2. **Recall the formula:** To rationalize denominators with cube roots, we use the identity for the sum of cubes: $$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$$ 3. **Set variables:** Let $$a = \sqrt[3]{5}, b = \sqrt[3]{7}, c = 2^{1/3}$$ but since 2 is rational, we keep it as is for now. Actually, since 2 is rational, we treat it as $$c = 2$$ (not cube root). So we want to rationalize $$a + b + c$$ where $$a = \sqrt[3]{5}, b = \sqrt[3]{7}, c = 2$$. 4. **Use the factorization:** Multiply numerator and denominator by $$a^2 + b^2 + c^2 - ab - bc - ca$$ to get: $$\frac{1}{a + b + c} \times \frac{a^2 + b^2 + c^2 - ab - bc - ca}{a^2 + b^2 + c^2 - ab - bc - ca} = \frac{a^2 + b^2 + c^2 - ab - bc - ca}{(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)}$$ 5. **Calculate the denominator:** Using the identity: $$(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) = a^3 + b^3 + c^3 - 3abc$$ 6. **Evaluate each term:** $$a^3 = 5, \quad b^3 = 7, \quad c^3 = 2^3 = 8$$ $$abc = \sqrt[3]{5} \times \sqrt[3]{7} \times 2 = 2 \times \sqrt[3]{35}$$ 7. **Substitute:** $$a^3 + b^3 + c^3 - 3abc = 5 + 7 + 8 - 3 \times 2 \times \sqrt[3]{35} = 20 - 6\sqrt[3]{35}$$ 8. **Write the rationalized form:** $$\frac{a^2 + b^2 + c^2 - ab - bc - ca}{20 - 6\sqrt[3]{35}}$$ 9. **Calculate numerator terms:** $$a^2 = (\sqrt[3]{5})^2 = \sqrt[3]{25}, \quad b^2 = \sqrt[3]{49}, \quad c^2 = 2^2 = 4$$ $$ab = \sqrt[3]{5} \times \sqrt[3]{7} = \sqrt[3]{35}, \quad bc = \sqrt[3]{7} \times 2 = 2\sqrt[3]{7}, \quad ca = 2 \times \sqrt[3]{5} = 2\sqrt[3]{5}$$ 10. **Final numerator:** $$\sqrt[3]{25} + \sqrt[3]{49} + 4 - \sqrt[3]{35} - 2\sqrt[3]{7} - 2\sqrt[3]{5}$$ **Answer:** $$\frac{\sqrt[3]{25} + \sqrt[3]{49} + 4 - \sqrt[3]{35} - 2\sqrt[3]{7} - 2\sqrt[3]{5}}{20 - 6\sqrt[3]{35}}$$