Attempt to solve #f(y) = -5y^3+6y^2-4 = 0#
First divide through by #-y^3# to get:
#5-6/y+4/y^3 = 0#
Let #x = 1/y#
Then #4x^3-6x+5 = 0#
Now let #x = u + v#
#0 = 4(u+v)^3 - 6(u+v) + 5#
#=4u^3+4v^3+(12uv-6)(u+v)+5#
#=4u^3+4v^3+6(2uv-1)(u+v)+5#
Let #v = 1/(2u)#
#=4u^3+1/(2u^3)+5#
Multiply through by #2u^3# to get:
#8(u^3)^2+10(u^3)+1 = 0#
#u^3 = (-10+-sqrt(100-32))/16#
#=(-10+-sqrt(68))/16#
#=(-5+-sqrt(17))/8#
Write:
#u_1 = root(3)((-5+sqrt(17))/8)#
#v_1 = root(3)((-5-sqrt(17))/8)#
Then real root of #4x^3-6x+5 = 0# is
#x = u_1 + v_1#
The other two (complex) roots are:
#x = omega u_1 + omega^2 v_1#
#x = omega^2 u_1 + omega v_1#
where #omega = -1/2+sqrt(3)/2i#
#y = 1/x#
So the real root of #f(y) = 0# is #y_1 = 1/(u_1+v_1)#
and the complex roots are:
#y_2 = 1/(omega u_1 + omega^2 v_1)#
#y_3 = 1/(omega^2 u_1 + omega v_1)#
#f(y) = -5(y - y_1)(y - y_2)(y - y_3)#
