All the variables are natural numbers. I'm not asking for a proof, since while we simply have$\ f(n)=n^3-n+1$,$\ g$ is a very long sum of cube roots (which contain square roots as well). I'm after general hints, or a general (kind of) procedure if it exists.
Given the sum$\ S$ here: https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-xfp1/v/t1.0-9/10702119_827019777360332_7596785284762545262_n.jpg?oh=da766c522af76997437a7904407a0824&oe=54B19B3E&gda=1425151927_aa94b35de563431507ee468820a8f7a2 ,
Let$$\ T= S-\sqrt[3]{\sqrt{\left(\frac{a_1a_2}{2}-\frac{5}{2}\right)^2-\frac{1}{27}}+\frac{a_1a_2}{2}-\frac{5}{2}} - \sqrt[3]{-\sqrt{\left(\frac{a_1a_2}{2}-\frac{5}{2}\right)^2-\frac{1}{27}}+\frac{a_1a_2}{2}-\frac{5}{2}} +\\ \sqrt[3]{\sqrt{\left(\frac{a_3a_4}{2}-\frac{137}{54}\right)^2-\frac{1}{729}}+\frac{a_3a_4}{2}-\frac{137}{54}}+\\\sqrt[3]{-\sqrt{\left(\frac{a_3a_4}{2}-\frac{137}{54}\right)^2-\frac{1}{729}}+\frac{a_3a_4}{2}-\frac{137}{54}}-\\ \sqrt[3]{\sqrt{\left(\frac{a_5a_6}{2}-\frac{133}{54}\right)^2-\frac{1}{729}}+\frac{a_4a_5}{2}-\frac{133}{54}} -\\\sqrt[3]{-\sqrt{\left(\frac{a_5a_6}{2}-\frac{133}{54}\right)^2-\frac{1}{729}}+\frac{a_4a_5}{2}-\frac{133}{54}}-1. \\ \\g=T^3-T+1$$
I think this will help us. I have found that for natural$\ n$ and real$\ x$,$$\ x^3-x+1=n \implies x= \sqrt[3]{\frac{n-1}{2}+ \frac{\sqrt{69+243(n-2)+81(n-2)(n-3)}}{18}} +\\\sqrt[3]{\frac{n-1}{2}- \frac{\sqrt{69+243(n-2)+81(n-2)(n-3)}}{18}}.$$ I shall note that the$\ m_i$ in the title just stand for $\ 28$ natural variables. Then in the photo I had already used other letters, so it would have been useless to use$\ m_i$ defining$\ T$.