Spotting that $\,x^8 + x^7 + 1\,$ is reducible.

The trick is that $8 \equiv 2 \pmod 3$ and $7 \equiv 1 \pmod 3.$ That means that, if we take a cube root of unity, say $$ \omega = \frac{-1 + i \sqrt 3}{2}, $$ we get $$ \omega^8 + \omega^7 + 1 = \omega^2 + \omega + 1 = 0. $$ The same holds for $$ \bar{\omega } = \omega^2. $$ That means that $$ (x - \omega)(x - \bar{\omega}) = x^2 + x + 1 $$ divides the original polynomial, as the quadratic is the minimal polynomial for $\omega.$

As you noticed, the same thing can be done with other roots of unity. If we had $$ x^{1347} + x^{949} + x^{216} + x^{98} + 1, $$ the exponents are $2,4,1,3 \pmod 5,$ and the displayed polynomial is divisible by $$ x^4 + x^3 + x^2 + x + 1 $$