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$x=\sqrt{3+2 \sqrt{2}}$

$x^2=3+2 \sqrt{2}$

$\dfrac{1}{x^2} = \dfrac{1}{3+2\sqrt{2}} = 3 - 2\sqrt{2}$ (Rationalizing the denominator)

$x^2 + \dfrac{1}{x^2} = 3+2 \sqrt{2} + 3-2 \sqrt{2} = 6$

$\left(x^2+ \dfrac{1}{x^2}\right)^2 = 6^2$

$x^4 + 2 + \dfrac{1}{x^4} = 36$

$\large{\boxed{x^4 + \dfrac{1}{x^4} = 34}}$

Given that, $x = \sqrt{3+2\sqrt{2}}$ .. Squaring both side, $x^2 = 3+2\sqrt{2}$ $\Rightarrow x^4 =17+12\sqrt{2}$ ..... $(1)$ Again, $\frac{1}{x^2} =3-2\sqrt{2}$ $\Rightarrow \frac{1}{x^4} = 17-2\sqrt{2}$ ........ $(2)$ Now, applying $(1)+ (2)$ \Rightarrow x^4 + \frac{1}{x^4}=\boxed{\color\red{34}}

$\displaystyle x = \sqrt { 3 + 2\sqrt { 2 } }, x ^ { 4 } + \frac { 1 } { x ^ { 4 } } = ( \sqrt {3 + 2 \sqrt { 2 } } ) ^ { 4 } + \frac { 1 } { ( \sqrt { 3 + 2 \sqrt { 2 } } ) ^ { 4 } } = \boxed { 34 }$ .