Question 983405
Table of common logarithms could work well.


{{{log((32.768^(1/3)))}}}
{{{(1/3)log((32.768))}}}
{{{(1/3)log((3.2768*10^1))}}}
{{{(1/3)(log((3.2768))+log((10^1)))}}}
{{{(1/3)(1+log((3.2768)))}}}
{{{(1/3)(1.51544)}}}
{{{0.50514}}}


Next, find antilog, base ten, of 0.50514, and find the final result to be near to 3.1999.