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.
