@Yemon Choi. You are correct. The numerical solution given by "Nullomolgous" does satisfy the equation. Also his equation look's like a general solution just as "Joe Silverman" commented. I was surprised that the equation is not primitive & has a common factor. Usually a general solutions do not have a common factor like the pythagoras equation of second degree. Namely the general solution, [(m^2+n^2),(m^2-n^2),(2mn)].

@Joe Silverman. Your comment about solution given by "Nullomolgous" being a general solution is incorrect because his equation does not produce the numerical solution, (x,y,z)=((17/56),(37/56),(3/4)). Also [z=3/4] in the equation cannot be represented as sum of two rational cubes.

@user44191. Thanks for catching my typo. You have a typo too. In your value's for (x,y,z) the variable 't ' need's to have the power two. There is a difference in your method & my method. I took the sum of two cubes [ (m^2-1)^3+1^3] . Hence we get it= m^2(m^4-3m^2+3). Thus we just need to multiply the LHS by (m^4-3m^2+3) to make the LHS a square.