f(x/4)= x^3+5
Too late, answer is 5+1/64(x^3)
no it's {x=(((sqrt(3)*i-1)*(f(((x)/4))-5)^((1/3)))/2), x=-((((sqrt(3)*i+1)*(f(((x)/4))-5)^((1/3)))/2)), x=(f(((x)/4))-5)^((1/3))}