$\newcommand{\R}{\mathbb R}$The roots are 
\begin{equation}	
\begin{aligned}
	x_1&:=\frac{1}{2} \left(m-\frac1{\sqrt3}\sqrt{\frac{4 \sqrt[3]{2} m^4}{t}-5 m^2+2^{5/3} t}\right), \\ 
	x_2&:=\frac{1}{2} \left(m+\frac1{\sqrt3}\sqrt{\frac{4 \sqrt[3]{2} m^4}{t}-5 m^2+2^{5/3} t}\right), \\ 
	x_3&:=\frac{1}{2} \left(m-\sqrt{m^2-4 \left(\frac{\left(1+i \sqrt{3}\right) m^4}{3\ 2^{2/3} t}+\frac{2
   m^2}{3}+\frac{\left(1-i \sqrt{3}\right) t}{6 \sqrt[3]{2}}\right)}\right), \\ 
	x_4&:=\frac{1}{2} \left(m+\sqrt{m^2-4 \left(\frac{\left(1+i \sqrt{3}\right) m^4}{3\ 2^{2/3} t}+\frac{2
   m^2}{3}+\frac{\left(1-i \sqrt{3}\right) t}{6 \sqrt[3]{2}}\right)}\right), \\ 
	x_5&:=\frac{1}{2} \left(m-\sqrt{m^2-4 \left(\frac{\left(1-i \sqrt{3}\right) m^4}{3\ 2^{2/3} t}+\frac{2
   m^2}{3}+\frac{\left(1+i \sqrt{3}\right) t}{6 \sqrt[3]{2}}\right)}\right), \\ 
	x_6&:=\frac{1}{2} \left(m+\sqrt{m^2-4 \left(\frac{\left(1-i \sqrt{3}\right) m^4}{3\ 2^{2/3} t}+\frac{2
   m^2}{3}+\frac{\left(1+i \sqrt{3}\right) t}{6 \sqrt[3]{2}}\right)}\right), 
\end{aligned}
\end{equation}
where 
\begin{equation}
	t:=\sqrt[3]{3 \sqrt{3} \sqrt{L \left(27 L-4 m^6\right)}-27 L+2 m^6}. 
\end{equation}

Clearly, $x_1\in\R\iff x_2\in\R$, $x_3\in\R\iff x_4\in\R$, and $x_5\in\R\iff x_6\in\R$, since $m$ is real. 

Using the Mathematica `Reduce` command, we get that 

 - $x_3$ and $x_5$ are never real and hence $x_4$ and $x_4$ are never real (if $m$ and $L$ are positive integers)
 
 - $x_1\in\R\iff x_2\in\R\iff m\ge x_2$, where $x_2$ is the positive root $x$ of the equation $9x^6=64L$. 
 
Thus, only $x_1$ and $x_2$ can ever be real roots of your sextic equation, and they are actually real (for given positive integers $m$ and $L$) iff $L\le\frac98\,m^6$.