1. **Problem:** Find the cube root of $1 \frac{127}{216}$. 2. **Step 1:** Convert the mixed fraction to an improper fraction: $$1 \frac{127}{216} = \frac{216}{216} + \frac{127}{216} = \frac{343}{216}$$ 3. **Step 2:** Use the cube root formula: $$\sqrt[3]{\frac{a}{b}} = \frac{\sqrt[3]{a}}{\sqrt[3]{b}}$$ 4. **Step 3:** Find cube roots of numerator and denominator: $$\sqrt[3]{343} = 7 \quad \text{(since }7^3=343\text{)}$$ $$\sqrt[3]{216} = 6 \quad \text{(since }6^3=216\text{)}$$ 5. **Step 4:** So, $$\sqrt[3]{\frac{343}{216}} = \frac{7}{6} = 1 \frac{1}{6}$$ 6. **Answer:** The cube root is $1 \frac{1}{6}$, which corresponds to option (C).