I have been given the polynomial $$f(x)=x^8+x^7-x^5-x^4-x^3+x+1$$ and have been asked to show it is irreducible over $\mathbb{Q}$ by considering the product $(x^2-x+1)f$. (Looking it up, I realise $f$ is in fact a cyclotomic polynomial itself (it's $\Phi_{30}$) so is irreducible straight away, but I can't just quote this - I am allowed to use the general fact that cyclotomic polynomials are irreducible over $\mathbb{Q}$, but have to actually prove it is cyclotomic before applying this).
My attempt so far: $x^2-x+1$ is the 6th cyclotomic polynomial, and can be written $(x-e^{\pi i/3})(x-e^{-\pi i/3})$ and is itself irreducible over $\mathbb{Q}$. The product $(x^2-x+1)f$ turns out to be $$x^{10}-x^5+1=\Phi_6(x^5)$$ but I am unsure what the next step is here - it seems important that this is effectively the same cyclotomic polynomial in a different variable. How is considering the product $(x^2-x+1)f$ meant to help?
You're on the right track. As you already know, $$x^{10}-x^5+1=(x^2-x+1)f(x)$$ Now, note that $$(e^{i\pi/15})^{10}-(e^{i\pi/15})^5+1=e^{2i\pi/3}-e^{i\pi/3}+1=e^{2i\pi/3}+e^{4i\pi/3}+1=0$$ Since $e^{i\pi/15}$ is a root of $x^{10}-x^5+1$, then it must be the root of at least one of the factors $x^2-x+1$ or $f(x)$. Note that $\Phi_{30}(x)$, the minimal polynomial of $e^{i\pi/15}$ in $\mathbb{Q}[x]$, is of degree $\varphi(30)=8$, so $e^{i\pi/15}$ is not a root of $x^2-x+1$, which means that $e^{i\pi/15}$ must be a root of $f(x)$. Since $\Phi_{30}(x)$ is the unique monic minimal polynomial of $e^{i\pi/15}$ in $\mathbb{Q}[x]$, the fact that $e^{i\pi/15}$ is a root of $f(x)$ and $\deg(f)=\deg(\Phi_{30})$ implies that $f(x)=\Phi_{30}(x)$. Finally, since $f(x)$ is cyclotomic, it is irreducible over $\mathbb{Q}$.