Square of Tangent of 67.5 Degrees

Note that $\dfrac{1}{x} = \dfrac{1}{(3 + 2\sqrt{2})}*\dfrac{3 - 2\sqrt{2}}{(3 - 2\sqrt{2})} = 3 - 2\sqrt{2}$ ,

and so $x + \dfrac{1}{x} = 6$ .

Next, observe that $x^{4} + \dfrac{1}{x^{4}} = (x + \dfrac{1}{x})^{4} - 4(x + \dfrac{1}{x})^{2} + 2$ .

We thus have that $x^{4} + \dfrac{1}{x^{4}} = 6^4 - 4*6^{2} + 2 = \boxed{1154}$ .

$\dfrac{1}{x}=\dfrac{1}{3+2\sqrt2.} =3-2\sqrt2\\\therefore~x + \dfrac{1}{x} =6 ~~\therefore~ (x + \dfrac{1}{x} )^2 = x^2 + 2 + \dfrac{1}{x^2} = 36~~ \implies~x^2 + \dfrac{1}{x^2} = 34\\ (x^2 + \dfrac{1}{x^2} )^2=x^4 + 2 + \dfrac{1}{x^4} =34^2.\\x^4 + \dfrac{1}{x^4}= 34^2-2= \boxed{ 1154 }$

To make 1/x^4 a surd, you change the sign, (3-2r2)^4

This means that when (3+2r2)^4 and (3-2r2)^4 are added, all negative values are cancelled

Using pascals triangle, you can deduce: 1,4,6,4,1 -- the fours are cancelled as they are negative values.

Therefore:

2.3^4 + 2.6.3^2.(2r2)^2 + 2.(2r2)^4

=162 + 864 + 128

=1154

observe that: $\frac{1}{x} = \frac{1}{3+2\sqrt{2}}\times\frac{3-2\sqrt{2}}{3-2\sqrt{2}}$ so using the identity $(a+b)^2 +(a-b)^2 = 2(a^2 +b^2) \Rightarrow\ x^2 + \dfrac{1}{x^2} = 2(9+8) = 34$ Now $x^4 + \dfrac{1}{x^4} = (x^2 +\dfrac{1}{x^2})^2 - 2 = 34^2 - 2 = \boxed{1154}$