that $x=0$ is a zero of $A(G;x)$ with multiplicity $n-{\Delta}$. The positivity of all coefficients of $A(G;x)$ gives $A(G;x)\neq 0$ for every $x>0$. Finally, by item vi), $A(G;x)=(-1)^{n+{\Delta}}A(G;-x)\neq0$ for every $x<0$.]{}

Note that Theorem \[t:PropRegular\] (items ii, iv, v and vi) has the following direct consequence.

\[c:Cubic\] Let $G$ be any cubic graph. Then, $$A(G;x)=A_{-3}(G)\, x^{n-3}+A_{-1}(G)\, x^{n-1}+A_1(G)\, x^{n+1}+A_{3}(G)\, x^{n+3},$$ with $A_{-3}(G)=n < m\le A_{-1}(G)$ and $A_{1}(G)\ge A_{3}(G)$.

A finite sequence of real numbers $(a_{0} , a_{1} , a_{2} , . . . , a_{n})$ is said to be *unimodal* if there is some $k \in \{0, 1, . . . , n\}$, called the *mode* of the sequence, such that $$a_{0} \leq . . . \leq a_{k-1} \leq a_{k} \geq a_{k+1} \geq . . . \geq a_{n}.$$ A polynomial is called unimodal if the sequence of its coefficients is unimodal. Therefore, we have the following result for alliance polynomial of a cubic graph, note that $A^\dag$ does not satisfy it.

\[c:CubUnim\] For any cubic graph $G$, $x^{(3-n)/2}\,A(G;\sqrt{x}\,)$ is an unimodal polynomial.

\[c:ConComp\] Let $G$ be any connected graph. Then $G$ is regular if and only if $A_{{\Delta}}(G) = 1$.

If $G$ is regular, then by Theorem \[p:AlliPoly\] vi) we obtain 