Your question involves arbitrary parameters $a, b, c \in \mathbf Z$, so the desired cyclicity does certainly not hold in general, as pointed out by @Dietrich Burde. I outline here a systematic - one could say algorithmic - study which involves easy but cumbersome calculations, so I suggest that you take a particular example (see e.g. Galois Group of $\sqrt{2+\sqrt{2}}$ over $\mathbb{Q}$) and check at every stage to have an idea of the method. Let $\alpha$ be as you say, and $K=\mathbf Q(\alpha)$. You want to show that $K/\mathbf Q$ is cyclic of degree 4. Proceed step by step:
1) It is clear that $K$ contains $L=\mathbf Q (\sqrt c)$ and $L=K(\sqrt\beta)$, with $\beta=a+b\sqrt c \in L$. In order that $[K:\mathbf Q]=4$, the two intermediate extensions must be quadratic, so you must check first that $c$ is not a square in $\mathbf Q^*$, then $\beta$ is not a square in $L^*$ (by identification)
2) Once you have checked 1), i.e. that $[K:\mathbf Q]=4$, there is a general procedure for showing that $K/\mathbf Q$ is Galois. Denote by $\sigma$ the $\mathbf Q$-automorphism of $L$ sending $\sqrt c$ to $-\sqrt c$. Then (easy Galois exercise) $K/\mathbf Q$ is Galois iff $\sigma(\beta) /\beta$ is a square in $L^*$ (check by identification)
3) Following 2), we know now that $K/\mathbf Q$ is Galois of degree 4. But a group $G$ of order 4 is abelian, bicyclic of type (2, 2) or cyclic of order 4. By Galois theory, $G$ is bicyclic iff $K$ is of the form $\mathbf Q (c, e)$, with $e\in \mathbf Q^*$ but $e\notin \mathbf Q^{*2}$ and $ec^-1\notin \mathbf Q^{*2}$, which amounts to (easy Galois exercise) $\beta$ satisfying $\beta e^{-1}\in L^{*2}$ (really tedious checking). Discard that case, and you get that $K/\mathbf Q$ is cyclic
3b) There is a less tedious way than the last part of 3). Following 2), we have checked that $\sigma(\beta) /\beta=x^2$, with $x\in L^*$. It follows that $N(x^2) =1$, or $N(x)=\pm 1$, where $N$ denotes the norm of $L/\mathbf Q$. Then $K/\mathbf Q$ is cyclic iff $N(x)=- 1$ (not difficult Galois exercise) ./.
