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 $$0\le u\le3,\quad v\le(u/3)^{3/2},$$ $$\begin{aligned}p(t_*)&=\frac{(4 u+3)^{3/2}}{3 \sqrt{15}}-u-2 v \\ &\ge\frac{(4 u+3)^{3/2}}{3 \sqrt{15}}-u-2 (u/3)^{3/2}\ge0, \end{aligned}$$ $$t_*>1/5\implies u<3/5,$$ $$\begin{aligned}p(1/5)&=\frac{1}{25} (5 u-50 v+9) \\ &\ge\frac{1}{25} (5 u-50(u/3)^{3/2}+9)\ge0\text{ if }0\le u<3/5. \end{aligned}$$ Also, $p(-\infty+)=-\infty<0$.
Thus, $p(t)=0$ for some real $t\le1/5$. $\quad\Box$