Theo định lý Viete: $\left\{\begin{matrix} x_{1}+x_{2}=2702 & \\x_{1}x_{2}=1 & \end{matrix}\right.$
Xét
$- (\sqrt{x_{1}}+\sqrt{x_{2}})^2=x_{1}+x_{2}+2\sqrt{x_{1}x_{2}}=2703$ $\Rightarrow \sqrt{x_{1}}+\sqrt{x_{2}}=\sqrt{2703}$
$- (\sqrt[3]{x_{1}}+\sqrt[3]{x_{2}})^3=x_{1}+x_{2}+3\sqrt[3]{x_{1}x_{2}}(\sqrt[3]{x_{1}}+\sqrt[3]{x_{2}})=2702+3(\sqrt[3]{x_{1}}+\sqrt[3]{x_{2}})$
Đặt $\sqrt[3]{x_{1}}+\sqrt[3]{x_{2}}=a$, giải pt $a^3-3a-2702=0$
⇔$(a - 14)(a²+ 14a +193) = 0$
⇔ $a = 14$
Khi đó: $M=14+\sqrt{2703}$