Methods for efficiently factoring the cubic polynomial $x^3 + 1$ $$x^3 + 1$$
factors as 
$$(x^2 - x + 1)(x + 1) .$$
It would have taken me a few minutes to identify this. What are the various approaches to determining rapidly that it is factorable, and factoring it?
 A: In this case, the Rational Root Theorem tells us that if $r$ is a rational root of $x^3 + 1$, then $r$ is either $+1$ or $-1$. Substituting shows that $-1$ is indeed a root (and that $+1$ is not), so $(x - (-1))$ is a factor of $x^3 + 1$. Applying polynomial long division gives the desired factorization:
$$\color{#bf0000}{\boxed{x^3 + 1 = (x + 1)(x^2 - x + 1)}}.$$
Note that on the other hand, the discriminant of the quadratic on the r.h.s. of the equation is $-3 < 0$, so it does not factor any further (over $\Bbb R$).
In general, any cubic polynomial $p(x)$ over $\Bbb R$ has a root, so in principle we can always factor it as a product
$$p(x) = (a x + b) (A x^2 + B x + C),$$
and we can attempt to solve this equation by distributing and comparing like coefficients to produce a quadratic system in $a, b, A, B, C$. For general cubic polynomials, however, the solutions of this system are extremely unpleasant. Essentially, in the cases in which they are not (at least, when the coefficients of $p$ are rational), the above rational root-finding method generally works.
Note too that substituting $x = -y$ in the above formula and rearranging gives the similar and occasionally useful formula
$$y^3 - 1 = (y - 1) (y^2 + y +1).$$
A: Here is one approach. 
Note that $$x^3+1=x^3+3x^2+3x+1 -3x^2-3x=(x+1)^3-3x(x+1)=(x+1)((x+1)^2-3x)=(x+1)(x^2-x+1)$$
Another approach would be noticing that $x=-1$ is a solution to $x^3+1=0$. Then you can divide $x^3+1$ by $x+1$ with polynomial long division. 
The third approach would be to set $x^3+1=(x+a)(x^2+bx+c)$ and solving the system of equations. 
A fourth approach would be using the formula for geometric series. 
The sum of the geometric series $1$, $-x$, $x^2$ is $\frac{1-(-x)^3}{1-(-x)}=\frac{x^3+1}{x+1}$. 
Multiplying $x+1$ on each side gives us that $(1-x+x^2)(1+x)=1+x^3$. 
A: If you notice that $x=-1$ is a root (this is reasonably easy to see by inspection and is something most people would look out for), then you can write
$x^3+1 = (x+1)(Ax^2 + Bx + C)$.
The coefficients $A$ and $C$ are easily determined, since the the only way to get an $x^3$ term is by multiplying $Ax^2$ by $x$, so this gives $A = 1$. In the same manner, $C$ is easily determined, since the only term independent of $x$ is $C$, so that $C = 1$. This means that you only need to find $B$.
If you multiply the rhs and look at the coefficient of $x$ you have $B + C = 0$ so that $B = -C = -1$. The final result is
$x^3+1 = (x+1)(x^2 - x + 1)$.
A: As Travis has noted, the rational roots theorem is probably the most powerful/easy method to trying to factor a polynomial of your type.
You have some polynomial:
$$ax^3+bx^2+cx+d$$
If there are any rational roots, they will be of the form
$$x=\pm\frac{Factor(a)}{Factor(d)}$$
For example:
$$x^3+1$$
$$x=\pm\frac{Factor(1)}{Factor(1)}=\pm\frac11=\pm1$$
This is the only possible rational root.  You have to actually check if it works, sadly.
$$x\ne1,x=-1$$
If you had $6x^3-3x^2-x-2$, then:
$$x=\pm\frac{Factor(6)}{Factor(2)}=\pm\frac{1,2,3,6}{1,2}=\pm\frac{1,2,3,6}{1},\pm\frac{1,2,3,6}{2}$$
After you attempt to see if any of the solutions actually works be substituting them back in for $x$, you find that $x=1$ is the only one that works.
Proceed to factor the polynomial to find the other non-rational roots.
I note that if you allow complex roots that are with many cube roots and other radicals, then all cubic polynomials and quartic polynomials are factorable.
When it comes down to it, group theory and things like discriments are most likely the best way to go if you have really hard polynomials to factor.
