やってみた
t=(a+(√[-g(a)]+√[-g(0)])^(2/3)+(√[-g(a)]-√[-g(0)])^(2/3))/2
大先生に見てもらうためにc=√[-g(a)],d=√[-g(0)]として式を短くする
a^3=c^2-d^2 なので
t=((c^2-d^2)^(1/3)+(c+d)^(2/3)+(c-d)^(2/3))/2
g(t)=2t^3-3at^2+g(0)
=2(((c^2-d^2)^(1/3)+(c+d)^(2/3)+(c-d)^(2/3))/2)^3-3(c^2-d^2)^(1/3)(((c^2-d^2)^(1/3)+(c+d)^(2/3)+(c-d)^(2/3))/2)^2-d^2
https://www.wolframalpha.com/input/?i=2%28%28%28c%5E2-d%5E2%29%5E%281%2F3%29%2B%28c%2Bd%29%5E%282%2F3%29%2B%28c-d%29%5E%282%2F3%29%29%2F2%29%5E3-3%28c%5E2-d%5E2%29%5E%281%2F3%29%28%28%28c%5E2-d%5E2%29%5E%281%2F3%29%2B%28c%2Bd%29%5E%282%2F3%29%2B%28c-d%29%5E%282%2F3%29%29%2F2%29%5E2-d%5E2&;lang=ja
c≧0,d≧0 なので g(t)=0