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Let \(x=\sqrt[3]{150}\Rightarrow f(x)=x^3-150=0,\ f'(x)=3x^2\).

Newton–Raphson: \(x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=x_n-\frac{x_n^3-150}{3x_n^2}\)

Iterative formula: \(x_{n+1}=\frac{2x_n^3+150}{3x_n^2}=\frac13\left(2x_n+\frac{150}{x_n^2}\right)\)

Using it (since \(5^3=125,\ 6^3=216\), take \(x_0=5\)): \(x_1=5.3333,\ x_2=5.3134,\ x_3=5.3133,\ x_4=5.3133\). Hence \(\sqrt[3]{150}\approx 5.3133\) (check: \(5.3133^3\approx150.00\)).

Doubtly AI · 25 Oct 2026 AI-verified · Oct 2026 Report
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