For f(x)=x³−37, $x_{n+1}=x_n-(x_n^3-37)/(3x_n^2)=(2x_n+37/x_n^2)/3$. Starting from 1 gives x₁=13. Then x₂=(26+37/169)/3=8.73964, or 8.74. Two iterations from this poor starting guess are not yet close to the actual cube root.
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