Number of Non-Trivial Solutions of $x^3+y^3+z^3+t^3=0$ $(\!\!\mod \, 5)$

Question

I want to find the number of non-trivial solutions of $$x^3+y^3+z^3+t^3=0 \quad (\!\!\!\!\!\mod \,5)$$ I understand that in mod 5, we can take cube roots, that is every element is the cube root of some other element, so does this mean that this is equivalent to finding the number of non-trivial solutions to $$x+y+z+t=0 \quad (\!\!\!\!\!\mod \,5 )$$ where $x,y,z,t$ range from $0$ to $4$. In this case, we only need to choose the first three variables, then the fourth one is determined, so is this $$5^3-1$$ number of solutions, where the minus one is the trivial solution of all zeroes. Is this the correct approach? I have a feeling that I've made some elementary mistake somewhere, it doesn't feel entirely correct. Any help is appreciated.