Xjyuk

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.

Over the complex numbers the difference of cubes factors as
$$a^3-b^3=(a-b)(a-omega b)(a-omega^2b),$$
where $omegainBbbC$ is a primitive cube root of unity, i.e. either
$$omega=-frac12+fracsqrt32i
qquadtext or qquad
omega=-frac12-fracsqrt32i.$$
This factorization is unique; each of these two primitive cube roots of unity is the square of the other. So if the difference of cubes factors over the real numbers, then it must factor as above. But the linear factors above are real if and only if $a=0$ or $b=0$.

So the only examples where the resulting quadratic is reducible over $BbbR$ are the trivial ones:
$$a^3-0^3=acdot(a^2)qquadtext and qquad 0^3-b^3=-bcdot(b^2).$$

Alternatively, without referring to complex numbers, instead it suffices to note that a real quadratic polynomial is irreducible if (and only if) its discriminant is negative. The discriminant of
$$a^2+ab+b^2,$$
viewed as a polynomial in either $a$ or $b$, is
$$b^2-4b^2=-3b^2qquadtext or qquad a^2-4a^2=-3a^2,$$
respectively. Either way, it is negative unless $b=0$ or $a=0$, respectively. So the resulting quadratic is reducible if and only if $a=0$ or $b=0$.

Let $x = a/b$. Then factoring $a^3 - b^3$ is essentially the same as factoring $x^3 -1$. That cubic has one real root, $1$. and factors as
$$
(x-1)(x^2+x+1)
$$
The other two roots are the complex roots of the quadratic factor - the other two cube roots of $1$:
$$
-frac12 pm fracsqrt32i.
$$
So the quadratic does not factor over the real numbers.

Let $$delta:=b^2-4ac.$$
We can factor the quadratic over $mathbb R$ if and only if $deltageq 0$.

In the example above,we have
$$a = 13,b = 15,c = 9$$
hence,
$$delta = 15^2-4times 13times 9=-243<0$$
so the quadratic isn't reducible over $mathbb R$.