# Show that $1-999x^{888}\in \mathbb{Q}[x]$ is irreducible

As the title says, I'm supposed to show $$1-999x^{888}\in \mathbb{Q}[x]$$ is irreducible.

In a previous part of the question I had to show $$x^{888} -999\in \mathbb{Q}[x]$$ was irreducible which I did using Eisenstein's criterion, but the same can't be done here and I can't see any way to use the previous result.

To see that the reducibility of a polynomial implies the reducibility of its reciprocal:

Consider the polynomial $$P(x)=\sum_{i=0}^da_ix^i=a_0+a_ax+\cdots +a_dx^d$$. Then the "reciprocal" polynomial $$Q(x)$$ is defined by $$Q(x)=\sum_{i=0}^d a_ix^{d-i}=a_d+a_{d-1}x+\cdots + a_0x^d$$.

That's the situation you have, with $$P(x)=1-999x^{888}$$.

We want to show that $$P(x)$$ reducible implies that $$Q(x)$$ is reducible as well (given what you have already shown, that will complete the proof).

The key point is to note that the reciprocal is given by $$Q(x)=x^dP\left( \frac 1x\right)$$

Suppose $$P(x)$$ factors as $$F_1(x)\times F_2(x)$$ with degree $$F_i$$ = $$d_i$$ with $$1, and $$d=d_1+d_2$$. Then $$Q(x)=x^{d_1+d_2}F_1\left( \frac 1x\right)F_2\left(\frac 1x\right)=\left(x^{d_1}F_1\left( \frac 1x\right)\right)\times \left(x^{d_2}F_2\left( \frac 1x\right)\right)$$

From which we deduce that $$Q(x)$$, the reciprocal of $$P(x)$$,factors as the product of the reciprocals of the factors of $$P(x)$$, and we are done.