If ${\mathrm{Sin}}^2\mathrm{\theta }+3\mathrm{Cos\theta }-2=0$ then ${\mathrm{Cos}}^3\mathrm{\theta }+{\mathrm{Sec}}^3\mathrm{\theta }=$

${\mathrm{Sin}}^2\mathrm{\theta }+3\mathrm{Cos\theta }-2=0$

$$\begin{gathered} 1-{\cos }^2\mathrm{\theta }+3 \mathrm{cos\theta }-2=0 \\ {\cos }^2\mathrm{\theta }-3 \mathrm{cos\theta }+1=0 \\ \mathrm{cos\theta }+\frac{1}{\mathrm{cos\theta }}=3 \\ {\cos }^3\mathrm{\theta }+{\sec }^3\mathrm{\theta }={(\mathrm{cos\theta }+\mathrm{sec\theta })}^3-3(\mathrm{cos\theta }+\mathrm{sec\theta }) \\ =27-9=18 \end{gathered}$$