If $f:R\to R$ define by $f\left(x\right)=x^3-1$ , then $f^{-1}\{-2,0,7\}$ is 

$f\left(x\right)=x^3-1=y\Rightarrow x=f^{-1}\left(y\right)$

$\Rightarrow x^3=y+1\Rightarrow x=(y+1{)}^{\frac{1}{3}}$

$\Rightarrow x=\sqrt[3]{(y+1)}=f^{-1}\left(y\right)$

$\Rightarrow f^{-1}\left(x\right)=\sqrt[3]{x+1}$

$\therefore f^{-1}(-2)=\sqrt[3]{-2+1}=\sqrt[3]{-1}=-1$

$\therefore f^{-1}\left(0\right)=\sqrt[3]{1}=1$

$\therefore f^{-1}\left(7\right)=\sqrt[3]{7+1}=2$

$\therefore f^{-1}(-2,0,7)=\{-1,1,2\}$