Write $a,b,c$ instead of $\alpha,\beta,\gamma$. Let $p(t)$ be the polynomial in question. Let
$$u:=\cos^2a+\cos^2b+\cos^2c,\quad v:=\cos a\,\cos b\,\cos c,$$
$$t_*:=\frac{1}{30} \left(15-\sqrt{15} \sqrt{4 u+3}\right).$$
Then
$$0\le u\le3,\quad v\le(u/3)^{3/2},$$
$$\begin{aligned}p(t_*)&=\frac{(4 u+3)^{3/2}}{3 \sqrt{15}}-u-2 v \\
&\ge\frac{(4 u+3)^{3/2}}{3 \sqrt{15}}-u-2 (u/3)^{3/2} \\
&=:g(u)\ge g(3)=0
\end{aligned}$$
(the latter inequality holds because $u=3$ is the only critical point of $g$ and $g(0)>0$),
$$t_*>1/5\implies u<3/5,$$
$$\begin{aligned}p(1/5)&=\frac{1}{25} (5 u-50 v+9) \\
&\ge\frac{1}{25} (5 u-50(u/3)^{3/2}+9) \\
&=:h(u)\ge\min(h(0),h(3/5)) \\
&>0 \text{ if }0\le u<3/5
\end{aligned}$$
(the 2nd inequality in the latter display holds because $h$ is concave).
So, $p(\min(t_*,1/5))\ge0$.
Also, $p(-\infty+)=-\infty<0$.
Thus, $p(t)=0$ for some real $t\le1/5$. $\quad\Box$