It's all about the roots

Let $\large t=\sqrt[3]{2-x}$ $\large\Rightarrow t^3=2-x\Rightarrow x-1=1-t^3$ Hence given equation simplifies to: $\large t=1- \sqrt{1-t^3}$ Taking 1 in LHS and squaring both sides to get: $\large \Rightarrow t^2-2t+\not1=\not1-t^3$ $\large \Rightarrow t(t+2)(t-1)=0$ $\large t=-2,0,1\Rightarrow t^3=-8,0,1$ $\Large x=2-t^3=10,2,1 \\ \Large \Rightarrow\alpha=1, \beta=2, \gamma=10\\ \small\color{#69047E}{{(\text{since }\alpha<\beta<\gamma)}}$ Plugging values of $\alpha,\beta,\gamma$ to get: $\large\color{#0C6AC7}{\frac{\alpha^2+\beta^2+\gamma^2}{\gamma^2-\beta^2-\alpha^2}}=\dfrac{105}{95}=\dfrac{21}{19}$ $\therefore m=21, n=19$ $\Large\color{forestgreen}{\lfloor\frac{(m.n)^{^{m-n}}}{m+n}\rfloor}=\lfloor\frac{(21.19)^{2}}{21+19}\rfloor~~~~~~~~~~~~~~~~~\\ ~~\large=\lfloor\frac{159201}{40}\rfloor=\lfloor3980.025\rfloor\\~~~~~~~~~~~~~~~~~~~~~~~\huge=\boxed{\color{#007fff}{3980}}$