My resolution $${\quad \quad \quad(2\sqrt[3]{2})^2 = h^2+(\sqrt[3]{2})^2 \\ \iff ( \sqrt[3]{16})^2= h^2 + \sqrt[3]{4} \\ \iff \sqrt[3]{256} = h^2 + \sqrt[3]{4}\\ \iff h^2 = \sqrt[3]{256}- \sqrt[3]{4}\\ \iff h = \sqrt{\sqrt[3]{256}- \sqrt[3]{4}} = \sqrt[6]{256}- \sqrt[6]{4}\\ \iff h = \sqrt[6]{64\times 4}- \sqrt[6]{4}\\ \iff h = 2\sqrt[6]{4}-\sqrt[6]{4} = \sqrt[6]{4}}$$
I've tried it over and over again but always get the same result. But in the textbook it says it should be $\sqrt[6]{108}$ .
On
Notice that we sort of have "like terms" on the left-hand side:
\begin{aligned} h^2 &= (2 \sqrt[3]{2})^2 - (\sqrt[3]{2})^2 \\ &= 4 \sqrt[3]{4} - \sqrt[3]{4} \\ &= 3 \sqrt[3]{4} \end{aligned}
If we want to write this as one root -- that is, without anything in front -- we have to "unsimplify" the $3$. In particular, it can helpful at this stage to use fractional exponents rather than roots, since the rules for combining them are more direct. But in order to add fractional exponents, we need a common denominator:
$$ 3 \sqrt[3]{4} = 3^{3/3} \cdot 4^{1/3} = (3^3 \cdot 4)^{1/3} = 108^{1/3}$$
Now since it was $h^2$ that equaled $108^{1/3}$, we have $h = \pm (108^{1/3})^{1/2} = \pm 108^{1/6} = \pm \sqrt[6]{108}$. Apparently, your book wants the positive answer :).
As pointed out in the comments, you are wrong in 6th line.
$$h=\sqrt{\sqrt[3]{256}-\sqrt[3]{4}}\ne\sqrt{\sqrt[3]{256}}-\sqrt{\sqrt[3]{4}}$$
The correct way to do this, $$h{=\sqrt{\sqrt[3]{256}-\sqrt[3]{4}}\\ =\sqrt{\sqrt[3]{4^3\cdot4}-\sqrt[3]{4}}\\ =\sqrt{4\sqrt[3]{4}-\sqrt[3]{4}}\\ =\sqrt{3\sqrt[3]{4}}\\ =\sqrt{\sqrt[3]{4\cdot3^3}}\\ =\sqrt{\sqrt[3]{108}}\\ =\sqrt[6]{108} \quad \square}$$
Or, simply you can do, $$h^2{=(2\sqrt[3]{2})^2 - (\sqrt[3]{2})^2\\ =\left(2\sqrt[3]{2} + \sqrt[3]{2}\right)(2\sqrt[3]{2} - \sqrt[3]{2})\\ =\left(3\sqrt[3]{2}\right)\cdot(\sqrt[3]{2}) \\=3(\sqrt[3]{2})^2\\}$$
$$\therefore h=\sqrt3\cdot\sqrt[3]{2}=\sqrt[6]{3^3\cdot2^2}=\sqrt[6]{108}$$