If only it's a power of 3

$\begin{aligned}&{\text{Using, }}\cos (3x) = 4{\cos ^3}x - 3\cos x{\text{ }} \Rightarrow {\text{ }}\cos (60^\circ ) = 4{\cos ^3}(20^\circ ) - 3\cos (20^\circ ) = \frac{1}{2} \\&{\text{Dividing with }}{\cos ^3}(20^\circ ){\text{ derives , }}\boxed{{{\sec }^3}(20^\circ ) = 8 - 6{{\sec }^2}(20^\circ )}. \\&{\text{Now, starting with the given expression, }}\\&{\text{ }}2\sqrt[3]{{3{{\sec }^2}(20^\circ ){{\sin }^2}(10^\circ )}} \\&= 2\sqrt[3]{{3{{\sec }^2}(20^\circ ) \cdot \frac{1}{2}\left( {1 - \cos (20^\circ )} \right)}} \\&= \sqrt[3]{{{2^3} \cdot \frac{3}{2}\left( {{{\sec }^2}(20^\circ ) - \sec (20^\circ )} \right)}} \\&= \sqrt[3]{{12{{\sec }^2}(20^\circ ) - 12\sec (20^\circ )}} \\&= \sqrt[3]{{{{\sec }^3}(20^\circ ) + 12{{\sec }^2}(20^\circ ) - 12\sec (20^\circ ) - {{\sec }^3}(20^\circ )}} \\&= \sqrt[3]{{\left( {8 - 6{{\sec }^2}(20^\circ )} \right) + 12{{\sec }^2}(20^\circ ) - 12\sec (20^\circ ) - {{\sec }^3}(20^\circ )}} \\&= \sqrt[3]{{8 - 12\sec (20^\circ ) + 6{{\sec }^2}(20^\circ ) - {{\sec }^3}(20^\circ )}} \\&= \sqrt[3]{{{2^3} - 3({2^2})\sec (20^\circ ) + 3(2){{\sec }^2}(20^\circ ) - {{\sec }^3}(20^\circ )}} \\&= \sqrt[3]{{{{\left( {2 - \sec 20^\circ } \right)}^3}}} \\&= 2 - \sec 20^\circ \\&{\text{Therefore, }}a = 2{\text{ , }}b = - 1{\text{ , }}\therefore a+b = \boxed{\boxed{ 1}} \\ \end{aligned}$