Simplify: $\;$ $\dfrac{\sqrt[3]{a + \sqrt{2 - a^2}} \sqrt[6]{1 - a \sqrt{2 - a^2}}}{\sqrt[3]{1 - a^2}}$, $\;\;$ $\left|a\right| < 1$
$\dfrac{\sqrt[3]{a + \sqrt{2 - a^2}} \sqrt[6]{1 - a \sqrt{2 - a^2}}}{\sqrt[3]{1 - a^2}}$
$= \dfrac{\left[a + \sqrt{2 - a^2}\right]^{\frac{1}{3}} \left[1 - a \sqrt{2 - a^2}\right]^{\frac{1}{6}}}{\left[1 - a^2\right]^{\frac{1}{3}}}$
$= \dfrac{\left[\left(a + \sqrt{2 - a^2}\right)^{\frac{1}{6}}\right]^2 \left(1 - a \sqrt{2 - a^2}\right)^{\frac{1}{6}}}{\left(1 - a^2\right)^{\frac{1}{3}}}$
$= \dfrac{\left[\left(a + \sqrt{2 - a^2}\right)^2\right]^{\frac{1}{6}} \left(1 - a \sqrt{2 - a^2}\right)^{\frac{1}{6}}}{\left(1 - a^2\right)^{\frac{1}{3}}}$
$= \dfrac{\left(a^2 + 2a \sqrt{2 - a^2} + 2 - a^2\right)^{\frac{1}{6}} \left(1 - a\sqrt{2 - a^2}\right)^{\frac{1}{6}}}{\left(1 - a^2\right)^{\frac{1}{3}}}$
$= \dfrac{2^{\frac{1}{6}} \left(1 + a \sqrt{2 - a^2}\right)^{\frac{1}{6}} \left(1 - a \sqrt{2 - a^2}\right)^{\frac{1}{6}}}{\left(1 - a^2\right)^{\frac{1}{3}}}$
$= \dfrac{2^{\frac{1}{6}} \left[\left(1 + a \sqrt{2 - a^2}\right) \left(1 - a \sqrt{2 - a^2}\right)\right]^{\frac{1}{6}}}{\left(1 - a^2\right)^{\frac{1}{3}}}$
$= \dfrac{2^{\frac{1}{6}} \left[1 - a^2 \left(2 - a^2\right)\right]^{\frac{1}{6}}}{\left(1 - a^2\right)^{\frac{1}{3}}}$
$= \dfrac{2^{\frac{1}{6}} \left[1 - 2a^2 + a^4\right]^{\frac{1}{6}}}{\left(1 - a^2\right)^{\frac{1}{3}}}$
$= \dfrac{2^{\frac{1}{6}} \left[\left(1 - a^2\right)^2\right]^{\frac{1}{6}}}{\left(1 - a^2\right)^{\frac{1}{3}}}$
$= \dfrac{2^{\frac{1}{6}} \left(1 - a^2\right)^{\frac{1}{3}}}{\left(1 - a^2\right)^{\frac{1}{3}}}$
$= 2^{\frac{1}{6}} = \sqrt[6]{2}$