#f(x) = x^3-3x^2-1#

graph{x^3-3x^2-1 [-20, 20, -10, 10]}

By the rational roots theorem, any rational roots of #f(x)# would be #+-1# and neither is a root of #f(x) = 0# :

#f(-1) = -1-3-1 = -5#

#f(1) = 1-3-1 = -3#.

I suspect there's a missing #+3x# term in the question, but I'll have a go at the question as it stands.

First, let #y = x-1#

#y^3 = (x-1)^3 = x^3-3x^2+3x-1#

#-3y = -3(x-1) = -3x+3#

#-3 = -3#

So: #y^3-3y-3 = x^3-3x^2-1 = f(x)#

Next, using Cardano's method, let #y = u+v#

#0 = f(x) = y^3-3y-3 = (u+v)^3-3(u+v)-3#

#=u^3+3u^2v+3uv^2+v^3-3(u+v)-3#

#=u^3+3(uv-1)(u+v)-3+v^3#

#=u^3-3+v^3# (if #v=1/u#)

#=u^3-3+1/u^3#

Multiply both ends by #u^3# to get:

#(u^3)^2-3(u^3)+1 = 0#

So #u^3 = (3+-sqrt(5))/2#

Since this derivation is symmetric in #u# and #v#, we can put

#u^3 = (3+sqrt(5))/2#

#v^3 = (3-sqrt(5))/2#

So there's a real root:

#y = u+v = root(3)((3+sqrt(5))/2) + root(3)((3-sqrt(5))/2)#

So

#x_1 = y+1 = 1+root(3)((3+sqrt(5))/2) + root(3)((3-sqrt(5))/2)#

The other two roots are complex (!) :

#x_2 = 1+omega root(3)((3+sqrt(5))/2) + omega^2 root(3)((3-sqrt(5))/2)#

#x_3 = 1+omega^2 root(3)((3+sqrt(5))/2) + omega root(3)((3-sqrt(5))/2)#

where #omega = -1/2+i sqrt(3)/2# is the primitive cube root of unity.

#f(x) = (x-x_1)(x-x_2)(x-x_3) = # something horribly complicated.

I'm pretty sure this is more complex than you're supposed to know about and my suspicion about the missing term in the question is justified.