Let m \((\not\equiv \pm 1 (mod 9))\) be cubefree, and let \(m=ab^ 2\) with a squarefree and \((a,b)=1\). It is shown that an algorithm for calculating the fundamental unit of \(K(m^{1/3})\) is as follows. Let \(1=j_ 1<j_ 2<...<j_ s\leq ab-1\) be all the solutions of \(x^ 3\equiv 1\) (mod ab), and let \(x_ 0\) run through the values of \(j_ i+nab\) \((i=1,...,s\); \(n=0,1,...)\) in ascending order. Let \(x_ 1=<x_ 0m^{-1/3}>\), \(x_ 2=<x_ 0(a^ 2b)^{-1/3}>\), where \(< >\) denotes the nearest integer. Then the fundamental unit is \(x_ 0+x_ 1m^{1/3}+x_ 2(a^ 2b)^{1/3}\) for the first triple for which \(x^ 3_ 0+mx^ 3_ 1+a^ 2bx^ 3_ 2-3abx_ 0x_ 1x_ 2=1.\)
A similar algorithm is given for the case \(m\equiv \pm 1 (mod 9)\). The method can be extended to other cubic fields with signature one, but then the need to find integral bases introduces a complication. Several results on the units in bicubic fields \(K(m^{1/3},n^{1/3})\), in relation to the units in their pure cubic subfields, are proved. Some tables of results, for various m,n, are given.