Finding the value of $\sqrt[4]{(4+\sqrt7)^{-1}}\sqrt{1+\sqrt7}$ with other approaches It is a problem from a timed exam,

What is the value of $\sqrt[4]{(4+\sqrt7)^{-1}}\sqrt{1+\sqrt7}$ ?
$1)1\qquad\qquad2)\sqrt[4]2\qquad\qquad3)2\qquad\qquad4)2\sqrt[4]2$

I solved it with two approaches.
First approach,
$$\sqrt[4]{\frac{1}{4+\sqrt7}}\times\sqrt{1+\sqrt7}=\sqrt[4]{\frac{2}{(1+\sqrt7)^2}}\times\sqrt{1+\sqrt7}=\sqrt[4]2$$
Second approach,
$$\sqrt[4]{\frac{1}{4+\sqrt7}}\times\sqrt{1+\sqrt7}=\sqrt[4]{\frac{1}{4+\sqrt7}}\times\sqrt[4]{8+2\sqrt7}=\sqrt[4]{\frac{2(4+\sqrt7)}{4+\sqrt7}}=\sqrt[4]2$$
I'm wondering is it possible to solve this problem with other efficient approaches?
 A: Personally I would have gone for $x^4=\dfrac{(1+\sqrt{7})^2}{4+\sqrt{7}}=\dfrac{8+2\sqrt{7}}{4+\sqrt{7}}=2$
In which you discover along the way that $(1+\sqrt{7})^2$ comes out perfect for a simplification, while I feel that in your approach it is prerequisite.
Also I do not have to carry on drawing all these outer roots symbols...
A: If we call $ \ u \ = \ 4 + \sqrt7 \ \ , $ then we have $ \ 1 + \sqrt7 \ = \ u - 3 \ $ and $ \ \frac{1}{u}  \ = \ \frac{4 - \sqrt7}{9} \ \ . $  Consequently,
$$\sqrt[4]{(4+\sqrt7)^{-1}} \ · \ \sqrt{1+\sqrt7} \ \ = \ \ \sqrt[4]{\frac{(u \ - \ 3)^2}{u}} \ \ = \ \ \sqrt[4]{ \  u \ - \ 6 \ + \ \frac{9}{u} \  }$$
$$ = \ \ \sqrt[4]{ \ (4 + \sqrt7) \ - \ 6 \ + \ 9· \left(\frac{4 - \sqrt7}{9} \right) \ }  \ \ = \ \ \sqrt[4]{ \  4   \ - \ 6 \ + \   4    \ } \ \ = \ \ \sqrt[4]{ 2 } \ \ . $$
A: Your methods are perfectly alright even for a timed examination.
In particular, I liked the second solution.
Another possible way to achieve the same would be rationalising the denominator inside the first radical.
