How to express $\frac{\frac{1}{2}}{\sqrt[4]{\frac{3}4}-\sqrt[4]{\frac{1}{4}}}$ into $\frac{1}{2}\sqrt{5+3\sqrt{3}+2\sqrt{12+7\sqrt{3}}}$ I am just wondering how to express $\frac{\frac{1}{2}}{\sqrt[4]{\frac{3}4}-\sqrt[4]{\frac{1}{4}}}$ into $\frac{1}{2}\sqrt{5+3\sqrt{3}+2\sqrt{12+7\sqrt{3}}}$ (This one is taken from Wolfram BTW)
Clearly, $\frac{\frac{1}{2}}{\sqrt[4]{\frac{3}4}-\sqrt[4]{\frac{1}{4}}}$ can be rewritten as $\frac{1}{2(\sqrt[4]{\frac{3}{4}}-\sqrt[4]{\frac{1}{4}})}$, but what should I do next?
I am a little inexperienced in manipulating surds, so some helps are very much appreciated
Thank you!
 A: Start without the irrelevant $1/2$ and note that $\sqrt[4]{4} = \sqrt{2}$, so: $$\frac{1}{\sqrt[4]{\frac{3}{4}}-\sqrt[4]{\frac{1}{4}}} = \frac{\sqrt{2}}{\sqrt[4]{3} -1}.$$ Now try to rationalize the denominator in the standard fashion (using $(x-y)\cdot(x+y) = x^2 - y^2$) by multiplying with $\frac{\sqrt[4]{3}+1}{\sqrt[4]{3}+1}$ to get
$$ \frac{\sqrt{2}}{\sqrt[4]{3} -1}\cdot\frac{\sqrt[4]{3}+1}{\sqrt[4]{3}+1} = \frac{\sqrt{2}(\sqrt[4]{3}+1)}{\sqrt{3} -1}.$$
That gets us half-way to a rationalized denominator, so finish the job by multiplying  with $\frac{\sqrt{3}+1}{\sqrt{3}+1}$:
$$\frac{\sqrt{2}(\sqrt[4]{3}+1)}{\sqrt{3} -1}\cdot\frac{\sqrt{3}+1}{\sqrt{3}+1}= \frac{\sqrt{2}(\sqrt{3}+1)(\sqrt[4]{3}+1)}{2}.$$
Now that we have a rational denominator, the basic idea is to rewrite the numerator as the square root of the square of its previous value, expand the squares, combine terms and see where we are:
$$\frac{\sqrt{2}(\sqrt{3}+1)(\sqrt[4]{3}+1)}{2} = \sqrt{\frac{2}{4}(\sqrt{3}+1)^2 (\sqrt[4]{3}+1)^2} = \sqrt{\frac{2}{4}(3+2\sqrt{3}+1)(\sqrt{3} +2\sqrt[4]{3}+1)} = \sqrt{\frac{2}{4}(10+6\sqrt{3} +4(2+\sqrt{3})\sqrt[4]{3})} = \sqrt{5 + 3\sqrt{3} + 2(2+\sqrt{3})\sqrt[4]{3}}.$$
You can see that already we have part of the required answer. We use the same trick of taking the square root of the square (thereby leaving everything the same) one more time on the last term to get rid of the pesky fourth root. We then expand the square and collect terms as usual:
$$\sqrt{5 + 3\sqrt{3} + 2(2+\sqrt{3})\sqrt[4]{3}} = \sqrt{5 + 3\sqrt{3} + 2\sqrt{(2+\sqrt{3})^2\sqrt{3}}}= \sqrt{5 + 3\sqrt{3} + 2\sqrt{(4+4\sqrt{3} +3)\sqrt{3}}}= \sqrt{5 + 3\sqrt{3} + 2\sqrt{4\sqrt{3} +12  +3\sqrt{3}}} = \sqrt{5 + 3\sqrt{3} + 2\sqrt{12 +7\sqrt{3}}}.$$
A: We need to prove that
$$(\sqrt[4]3-1)^2(5+3\sqrt3+2\sqrt[4]3(2+\sqrt3))=2$$ or
$$2(5+3\sqrt3+2\sqrt[4]3(2+\sqrt3))=(1+\sqrt[4]3+\sqrt3+\sqrt[4]{27})^2,$$
which is obvious after full expanding.
A: You will probably have to use the identity: 
$$\dfrac{1}{x-y} = \dfrac{(x+y)(x^2+y^2)}{x^4-y^4} = \dfrac{x^3+xy^2+x^2y+y^3}{x^4-y^4}$$
