Further factorisation of a difference of cubes?

Over the complex numbers the difference of cubes factors as $$a^3-b^3=(a-b)(a-\omega b)(a-\omega^2b),$$ where $\omega\in\Bbb{C}$ is a primitive cube root of unity, i.e. either $$\omega=-\frac12+\frac{\sqrt{3}}{2}i \qquad\text{ or }\qquad \omega=-\frac12-\frac{\sqrt{3}}{2}i.$$ 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 $\Bbb{R}$ are the trivial ones: $$a^3-0^3=a\cdot(a^2)\qquad\text{ and }\qquad 0^3-b^3=-b\cdot(b^2).$$

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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 $$b^2-4b^2=-3b^2\qquad\text{ or }\qquad a^2-4a^2=-3a^2,$$ 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 $1$: $$ -\frac{1}{2} \pm \frac{\sqrt{3}}{2}i. $$ So the quadratic does not factor over the real numbers.