1. **State the problem:** We want to find an approximation to the cube root of 7, i.e., find $x$ such that $x^3 = 7$. 2. **Define the function:** Let $f(x) = x^3 - 7$. We want to find the root of $f(x)$, i.e., solve $f(x) = 0$. 3. **Newton-Raphson formula:** The iterative formula is $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$ where $f'(x) = 3x^2$ is the derivative of $f(x)$. 4. **Start with initial guess:** $x_0 = 1.3$. 5. **Iteration 1:** $$f(1.3) = 1.3^3 - 7 = 2.197 - 7 = -4.803$$ $$f'(1.3) = 3 \times 1.3^2 = 3 \times 1.69 = 5.07$$ $$x_1 = 1.3 - \frac{-4.803}{5.07} = 1.3 + 0.947 = 2.247$$ 6. **Iteration 2:** $$f(2.247) = 2.247^3 - 7 = 11.35 - 7 = 4.35$$ $$f'(2.247) = 3 \times 2.247^2 = 3 \times 5.05 = 15.15$$ $$x_2 = 2.247 - \frac{4.35}{15.15} = 2.247 - 0.287 = 1.960$$ 7. **Iteration 3:** $$f(1.960) = 1.960^3 - 7 = 7.54 - 7 = 0.54$$ $$f'(1.960) = 3 \times 1.960^2 = 3 \times 3.84 = 11.52$$ $$x_3 = 1.960 - \frac{0.54}{11.52} = 1.960 - 0.047 = 1.913$$ 8. **Iteration 4:** $$f(1.913) = 1.913^3 - 7 = 6.99 - 7 = -0.01$$ $$f'(1.913) = 3 \times 1.913^2 = 3 \times 3.66 = 10.98$$ $$x_4 = 1.913 - \frac{-0.01}{10.98} = 1.913 + 0.00091 = 1.914$$ 9. **Iteration 5:** $$f(1.914) = 1.914^3 - 7 \approx 7.0000 - 7 = 0.0000$$ $$x_5 \approx 1.914$$ 10. **Final answer:** The cube root of 7 is approximately $$\boxed{1.913997}$$ correct to six decimal places.