Hello again; this two are the last probs for some time. As usual, please check my work.

The directions are: Rationalize the denominator.

$\frac{1\,-\,\sqrt{2}}{2\sqrt{3}\,-\,\sqrt{6}}$

So I multilply by the conjugate & simplify: $\frac{\sqrt{6}\,-\,2\sqrt{6}}{12\,-\,6}\;=\boxed{\;-\,\frac{\,\sqrt{6}}{6}}$

Now this beastly problem w/the same directions: $\sqrt[3]{\frac{16}{9}}$

I thought to write it like this: $\frac{\sqrt[3]{16}}{\sqrt[3]{9}}$

Then multiply the top and bottom by $\sqrt[3]{3}$: $\frac{\sqrt[3]{48}}{3}\,=\,\boxed{\frac{2\sqrt[3]{6}}{3}}$

Does this look good? If you have another way that might be better, do say. Thanks once again!

2. [size=3]Hello, Jonboy!

Rationalize the denominator: . $\frac{1\,-\,\sqrt{2}}{2\sqrt{3}\,-\,\sqrt{6}}$

So I multilply by the conjugate & simplify: $\frac{\sqrt{6}\,-\,2\sqrt{6}}{12\,-\,6}\;=\boxed{\;-\,\frac{\,\sqrt{6}}{6}}$ . . . . Right!

Now this beastly problem w/the same directions: $\sqrt[3]{\frac{16}{9}}$

I thought to write it like this: $\frac{\sqrt[3]{16}}{\sqrt[3]{9}}$

Then multiply the top and bottom by $\sqrt[3]{3}$: $\frac{\sqrt[3]{48}}{3}\,=\,\boxed{\frac{2\sqrt[3]{6}}{3}}$ . . . . Great!
You did this one in the most efficient way possible.

Many (most?) would multiply top and bottom by $\sqrt[3]{81}$
. . I did ... many years ago.

3. Thanks for the confirmation Soroban! Also for posting that reply a pretty long time ago showing the different ways to rationalize the denominator, that's where I learned my shortcut.