Do it without calculator

Solution:

On the numerator we will use this formula:

I don't know the name of that identity in English, so if anyone knows, please, write a reply or something, that would mean a lot for me. I am learning mathematical terms.

One of the condition for using the identity is $a \ge \sqrt{b}$ , and that is fulfilled since $49 > 48$ . After that we will get that the numerator is as same as the denominator, therefore the fraction is $1$ .

We desire to write $7+4\sqrt{3}$ as a perfect square. This is done by saying $(a+b\sqrt{3})^2 = 7+4\sqrt{3}$ . We then have $a^2+3b^2 = 7$ and $ab =2$ . We can guess and check the solution $a = 2, b = 1$ . Then we see that $\sqrt{7+4\sqrt{3}} = 2+\sqrt{3}$ and $\left(\dfrac{\sqrt{7+4\sqrt{3}}}{2+\sqrt{3}}\right)^{2016} = 1$ .

Simply read the title of the problem and look at the equation with the power of $2016$ . The only possible way (that wouldn't take ages) to count it would be if the fraction inside was equal to either $1$ or $0$ or $-1$ and either way the answer would be 1. ;)

Notice that, $2 + \sqrt{ 3 } = \left| 2 + \sqrt{ 3 } \right| = \sqrt{ \left( 2 + \sqrt{ 3 } \right)^2 } = \sqrt{ 2^2 + 2 \times 2 \cdot \sqrt{ 3 } + \sqrt{ 3 }^2 } = \sqrt{ 7 + 4 \sqrt{ 3 } }$

Hence the expression simplifies to $\boxed{1}$

$7+4\sqrt{3}$ can be written as $4+3+2×2\sqrt{3}$ .

ie. $2^2+{(\sqrt{3})}^2+2×2\sqrt{3}$

or ${(2+\sqrt{3})}^{2}$

Thus our problem reduces to ${\dfrac{\sqrt{{(2+\sqrt{3})}^2}}{2+\sqrt{3}}}^{2016}$

Which is equal to $1^{2016}$ ie. $1$ .

Hence our answer is $1$ .