Context is a course covering field theory (ch. 13) and Galois theory (ch. 14) of Abstract Algebra by Dummit & Foote. I've showed (a), that $h(x)$ is irreducible in $\mathbb{Q}[x]$. A similar solution can be found in A curious condition for f being irreducible in Q[x]
(a) Let $p$ be a prime number. For a polynomial $F ∈ \mathbb{Z} [x]$ we denote by $\bar{F}$ the reduction of $F$ modulo $p$. Recall that this means that $\bar{F}$ is the polynomial in $\mathbb{F}_p[x]$ that arises by reducing the coefficients of $F$ modulo $p$. The reduction modulo $p$ gives a ring homomorphism $\mathbb{Z} [x] → \mathbb{F}_ p [x]$.
Now suppose that $n ∈ \mathbb{N}$ and $f, g ∈ \mathbb{Z} [x]$ are such that: $f$ is monic with $\bar{f}$ irreducible in $\mathbb{F}_p [x]$, $\bar{f}$ and $\bar{g}$ are relatively prime in $\mathbb{F}_ p [x]$ (meaning that $\bar{f}$ and $\bar{g}$ have no common factor of degree $> 0$ in $\mathbb{F}_p [x]$), and for the polynomial $h(x) := f(x)^n + p · g(x)$ we have deg $g$ $<$ deg $h$.
Show that then $h(x)$ is irreducible in $\mathbb{Q} [x]$. (Hint: Recall that, by Gauss’ lemma, if $h(x)$ is reducible in $\mathbb{Q} [x]$ then it is also reducible in $\mathbb{Z} [x]$.)
(b) Use part (a) to show the following. If $p$ is a prime number, $a ∈ \mathbb{Z}$ , $n ∈ \mathbb{N}$ , and $g ∈ \mathbb{Z} [x]$ has deg $g < n$, and is such that $p \nmid g(a)$ then the polynomial $(x − a)^n + p · g(x)$ is irreducible in $\mathbb{Q} [x]$.
(c) Let $p$ be a prime number. Use part (b) to give a proof – different from the one you can find in the book (page 554) – that the polynomial $Φ_p (x) := x^{p−1}+ . . . + x + 1$ is irreducible in $\mathbb{Q} [x]$. (Hint: Note that $Φ_p (x) =\frac{x^p−1}{x−1}$. Use the rule $(a+b)^p = a^p +b^p$ (“Freshman’s dream”) that holds in any ring of characteristic $p$; in particular, the rule holds in the polynomial ring $\mathbb{F}_p [x]$.)
So I'm at (b) and I suppose since $(x-a)$ is irreducible in $\mathbb{Z}[x]$, $\overline{(x-a)}$ is irreducible in $\mathbb{F}_p[x]$, so what is left to show is $\overline{(x-a)}$ and $\overline{g}$ are relatively prime before you can use the result of (a). I suppose it has something to do with $p \nmid g(a)$ but I don't know. No idea how to get started with (c).
(b) $\overline{g(a)} = \overline{g}(\overline{a})$. As $p \nmid g(a)$ we have $0 \neq \overline{g}(\overline{a}) \in \mathbb{F}_p$ , i.e. $\overline{a}$ is not a root of $\overline{g}$. This shows that $x − \overline{a}$ and $\overline{g}$ are relatively prime in $\mathbb{F}_p[x]$. As the reduction modulo $p$ of $x − a$ is $x − \overline{a}$, part (a) now applies.
(c) The polynomials $x^p − 1$ and $(x − 1)^p$ are equal in $\mathbb{F}_p [x]$. As a consequence, the reductions modulo $p$ of the polynomials $Φ_p(x)=\frac{x^p-1}{x-1}$ and $\frac{(x-1)^p}{x-1}=(x-1)^{p-1}$ are the same. This means $Φ_p(x)=(x-1)^{p-1}+p\cdot g(x)$ for some $g \in \mathbb{Z}[x]$. Since $Φ_p(x)$ and $(x − 1)^{p−1}$ are both monic, we must have deg $g$ $<$ deg $Φ_p = p − 1$. Since $Φ_p (1) = p$ we see $g(1) = 1$ and hence part (b) applies with $a = 1$.