Apply extend Bunyakovsky inequality ,we have:

\(a^3+b^3+c^3\)\(=\dfrac{a^4}{a}+\dfrac{b^4}{b}+\dfrac{c^4}{c}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a+b+c}=\dfrac{81}{a+b+c}\)                                              (1)                                      

Other way:

\(3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\Rightarrow27\ge\left(a+b+c\right)^2\)\(\Rightarrow a+b+c\le\sqrt{27}=3\sqrt{3}\)                 (2)

From (1) and (2)\(\Rightarrow a^3+b^3+c^3\ge\dfrac{81}{a+b+c}\ge\dfrac{81}{3\sqrt{3}}=9\sqrt{3}\)

\(\Rightarrow min_M=9\sqrt{3}\) when a=b=c=\(\sqrt{3}\) because a,b,c\(\ge-1\)