f(x) = x^3-3x^2-1f(x)=x3−3x2−1
graph{x^3-3x^2-1 [-20, 20, -10, 10]}
By the rational roots theorem, any rational roots of f(x)f(x) would be +-1±1 and neither is a root of f(x) = 0f(x)=0 :
f(-1) = -1-3-1 = -5f(−1)=−1−3−1=−5
f(1) = 1-3-1 = -3f(1)=1−3−1=−3.
I suspect there's a missing +3x+3x term in the question, but I'll have a go at the question as it stands.
First, let y = x-1y=x−1
y^3 = (x-1)^3 = x^3-3x^2+3x-1y3=(x−1)3=x3−3x2+3x−1
-3y = -3(x-1) = -3x+3−3y=−3(x−1)=−3x+3
-3 = -3−3=−3
So: y^3-3y-3 = x^3-3x^2-1 = f(x)y3−3y−3=x3−3x2−1=f(x)
Next, using Cardano's method, let y = u+vy=u+v
0 = f(x) = y^3-3y-3 = (u+v)^3-3(u+v)-30=f(x)=y3−3y−3=(u+v)3−3(u+v)−3
=u^3+3u^2v+3uv^2+v^3-3(u+v)-3=u3+3u2v+3uv2+v3−3(u+v)−3
=u^3+3(uv-1)(u+v)-3+v^3=u3+3(uv−1)(u+v)−3+v3
=u^3-3+v^3=u3−3+v3 (if v=1/uv=1u)
=u^3-3+1/u^3=u3−3+1u3
Multiply both ends by u^3u3 to get:
(u^3)^2-3(u^3)+1 = 0(u3)2−3(u3)+1=0
So u^3 = (3+-sqrt(5))/2u3=3±√52
Since this derivation is symmetric in uu and vv, we can put
u^3 = (3+sqrt(5))/2u3=3+√52
v^3 = (3-sqrt(5))/2v3=3−√52
So there's a real root:
y = u+v = root(3)((3+sqrt(5))/2) + root(3)((3-sqrt(5))/2)y=u+v=3√3+√52+3√3−√52
So
x_1 = y+1 = 1+root(3)((3+sqrt(5))/2) + root(3)((3-sqrt(5))/2)x1=y+1=1+3√3+√52+3√3−√52
The other two roots are complex (!) :
x_2 = 1+omega root(3)((3+sqrt(5))/2) + omega^2 root(3)((3-sqrt(5))/2)x2=1+ω3√3+√52+ω23√3−√52
x_3 = 1+omega^2 root(3)((3+sqrt(5))/2) + omega root(3)((3-sqrt(5))/2)x3=1+ω23√3+√52+ω3√3−√52
where omega = -1/2+i sqrt(3)/2ω=−12+i√32 is the primitive cube root of unity.
f(x) = (x-x_1)(x-x_2)(x-x_3) = f(x)=(x−x1)(x−x2)(x−x3)= 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.
