<s>[INST] How do i solve the cubic equation? 
$$x^3 - 3x^2 - 3x +2 = 0$$

The rational root test does not work; there are no rational roots. 
 [/INST] Here is a step by step procedure.
First shift $x$ so there is no quadratic term.  To do this we shift by 1/3 the quadratic term (assuming leading coefficient is 1) This give
$$
x=y+1 \Rightarrow y^3-6\,y-3 =0
$$
We now reduce the cubic by the substitution $y = A(z+1/z)$
$$
y = A(z+1/z) \Rightarrow \frac{z^6\,A^3+3\,z^4\,A^3+3\,z^2\,A^3+A^3-6\,z^4\,A-6\,z^2\,A-3\,z^3}{z^3}=0
$$
We now pick $A \neq 0$ so that $Z^4$ term is zero. This requires
$$ A(A^2-2)=0$$
So pick $A=\sqrt{2}$ and the equation becomes
$$2^{{{3}\over{2}}}\,z^6-3\,z^3+2^{{{3}\over{2}}}=0$$
This is a quadratic in $z^3$. So set $z=u^{1/3}$ to get
$$
z=u^{1/3} \Rightarrow 2^{{{3}\over{2}}}\,u^2-3\,u+2^{{{3}\over{2}}}=0$$
Solve this quadratic for $u$. You will have two roots.
Use them to find $z$. Recall each number will have 3 cube roots, but you will have only 3 possible values and not 6.
Once you have $z$, calculate $y$ and then finally $x$.
</s>