We want integers $ a $ and $ b $ so that \[a + b \sec 20^\circ = 2 \sqrt[3]{3 \sec^2 20^\circ \sin^2 10^\circ}.\]Cubing both sides, we get \[a^3 + 3a^2 b \sec 20^\circ + 3ab^2 \sec^2 20^\circ + b^3 \sec^3 20^\circ = 24 \sec^2 20^\circ \sin^2 10^\circ.\]From the half-angle formula, $ \sin^2 10^\circ = \frac{1 - \cos 20^\circ}{2}, $ so \begin{align*} 24 \sec^2 20^\circ \sin^2 10^\circ &= 24 \sec^2 20^\circ \cdot \frac{1 - \cos 20^\circ}{2} \\ &= 12 \sec^2 20^\circ - 12 \sec 20^\circ.\end{align*}To deal with the $ \sec^3 20^\circ $ term, we apply the triple angle formula $ \cos 3x = 4 \cos^3 x - 3 \cos x $. Setting $ x = 20^\circ, $ we get \[\frac{1}{2} = \cos 60^\circ = 4 \cos^3 20^\circ - 3 \cos 20^\circ.\]Dividing both sides by $ \cos^3 20^\circ, $ we get $ \frac{1}{2} \sec^3 20^\circ = 4 - 3 \sec^2 20^\circ, $ so \[\sec^3 20^\circ = 8 - 6 \sec^2 20^\circ.\]Thus, \begin{align*} &a^3 + 3a^2 b \sec 20^\circ + 3ab^2 \sec^2 20^\circ + b^3 \sec^3 20^\circ \\ &= a^3 + 3a^2 b \sec 20^\circ + 3ab^2 \sec^2 20^\circ + b^3 (8 - 6 \sec^2 20^\circ) \\ &= a^3 + 8b^3 + 3a^2 b \sec 20^\circ + (3ab^2 - 6b^3) \sec^2 20^\circ.\end{align*}We want this to equal $ 12 \sec^2 20^\circ - 12 \sec 20^\circ, $ so we can try to find integers $ a $ and $ b $ so that \begin{align*} a^3 + 8b^3 &= 0, \\ 3a^2 b &= -12, \\ 3ab^2 - 6b^3 &= 12.\end{align*}From the first equation, $ a^3 = -8b^3, $ so $ a = -2b $. Substituting into the second equation, we get $ 12b^3 = -12, $ so $ b^3 = -1, $ and $ b = -1 $. Then $ a = -2 $. These values satisfy the third equation, so $ (a,b) = \boxed{(2,-1)} $.