$\begin{align}  & x\sqrt{2+x}+2\sqrt{2-x}=\sqrt{8+{{x}^{3}}} \\ & \text{Square both sides} \\ & {{\left( x\sqrt{2+x}+2\sqrt{2-x} \right)}^{2}}={{\left( \sqrt{8+{{x}^{3}}} \right)}^{2}} \\ & {{x}^{2}}(2+x)+4x\sqrt{2-{{x}^{2}}}+4(2-x)=8+{{x}^{3}} \\ & 2{{x}^{2}}+{{x}^{3}}+8-4x-8-{{x}^{3}}=-4x\sqrt{4-{{x}^{2}}} \\ & 2{{x}^{2}}-4x=-4x\sqrt{4-{{x}^{2}}} \\ & \text{Divide through by }2x \\ & x-2=-2\sqrt{4-{{x}^{2}}} \\ & \text{Square both sides} \\ & {{x}^{2}}-4x+4=-4(4-{{x}^{2}}) \\ & {{x}^{2}}-4x+4=16-4{{x}^{2}} \\ & 5{{x}^{2}}-4x-12=0 \\ & 5{{x}^{2}}-10x+6x-12=0 \\ & 5x(x-2)+6(x-2)=0 \\ & (5x+6)(x-2)=0 \\ & x=-\frac{6}{5}\text{ or }x=2 \\ & \text{So by back}-\text{Substitution we see that }x=2\text{ and }-\frac{6}{5} \\\end{align}$