#2**+5 **

There is, I vaguely remember, an algebraic method similar to that for calculating square roots and it's also possible to use the Newton-Raphson method, (though you would need to be good at arithmetic, hardly the sort of calculation you would attempt without a calculator).

If the number is close to some convenient cube, it's possible to make use of the binomial expansion. For example.

$$\begin{array}{lcl}

\sqrt[3]{1005}&=&(1000+5)^{1/3}\\

&=&1000^{1/3}(1+0.005)^{1/3}\\

&=&10(1+(1/3)0.005+(1/3)(-2/3)0.005^{2}/2!+\dots)\\

&=&10(1+0.0016667-0.0000028+\dots)\\

&\approx&10.016639

\end{array}$$

That's easily done on paper and is correct to 6dp.

Bertie
May 30, 2014