Further factorisation of a difference of cubes? We know that a difference of cubes can be factored using $a^3-b^3=(a-b)(a^2+ab+b^2)$. How do we know that the quadratic can't be factored further.
For example, 
$$
27x^3-(x+3)^3=((3x-(x+3))(9x^2+3x(x+3)+(x+3)) \\
=(2x-3)(13x^2+15x+9)
$$
In this case we can't factor the quadratic over R. Is there any way to know before factorising the cubic that the resulting quadratic can't be factored further.
Another way of asking is to give an example similar to the one above where the resulting quadratic is reducible over R.

Let $$\delta:=b^2-4ac.$$
We can factor the quadratic over $\mathbb R$ if and only if $\delta\geq 0$.
In the example above,we have 
$$a = 13,b = 15,c = 9$$
hence,
$$\delta = 15^2-4\times 13\times 9=-243<0$$
so the quadratic isn't reducible over $\mathbb R$.
