f(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) 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
y3=(x−1)3=x3−3x2+3x−1
−3y=−3(x−1)=−3x+3
−3=−3
So: y3−3y−3=x3−3x2−1=f(x)
Next, using Cardano's method, let y=u+v
0=f(x)=y3−3y−3=(u+v)3−3(u+v)−3
=u3+3u2v+3uv2+v3−3(u+v)−3
=u3+3(uv−1)(u+v)−3+v3
=u3−3+v3 (if v=1u)
=u3−3+1u3
Multiply both ends by u3 to get:
(u3)2−3(u3)+1=0
So u3=3±√52
Since this derivation is symmetric in u and v, we can put
u3=3+√52
v3=3−√52
So there's a real root:
y=u+v=3√3+√52+3√3−√52
So
x1=y+1=1+3√3+√52+3√3−√52
The other two roots are complex (!) :
x2=1+ω3√3+√52+ω23√3−√52
x3=1+ω23√3+√52+ω3√3−√52
where ω=−12+i√32 is the primitive cube root of unity.
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.
