Let  $x*y=x^2+y^3$ and  $(x*1)*1=x*(1*1)$ then  $5{\sin }^{-1}\left(\frac{x^4+x^2-2}{x^4+x^2+2}\right)$ =____

$$\begin{gathered} x*y=x^2+y^3 \\ x*y*1=(x^2+1)*1 \\ ={(x^2+1)}^2+1 \\ =x^4+2x^2+2 \\ x*(1*1)=x*(1+1)=x*2=x^2+8 \\ \therefore x^4+2x^2+2=x^2+8 \\ {x}^4+x^2=6 \\ 5{\sin }^{-1}\left(\frac{6-2}{6+2}\right)=5{\sin }^{-1}\left(\frac{1}{2}\right)=5\frac{\pi }{6} \end{gathered}$$