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$(x^2 + 2)^2 - 6x(x^2 + 2) + 9x^2 - x^2 = 0$

$( x^2 + 2 - 3x)^2 - x^2 =0$

$(x^2 - 2x + 2)(x^2 - 4x + 2)=0$

$((x - 1)^2 + 1)((x - 2)^2 - 2) = 0$

Real solutions - $(x - 2)^2 = 2$

$x = 2 \pm \sqrt{2}$

difference between them = $2\sqrt{2} = \sqrt{8}$

$(x^2+2)^2+8x^2=6x(x^2+2) \quad \Rightarrow x^4+4x^2+4+8x^2 = 6x^3+12x$

$\Rightarrow x^4-6x^3+12x^2-12x+4 = 0\quad \Rightarrow x^2-6x+12-\dfrac {12}{x} + \dfrac {4} {x^2} = 0$

$\Rightarrow \left( x^2+4+\dfrac {4} {x^2} \right) - 6 \left( x+\dfrac {2} {x} \right) +12 -4 = 0$

$\Rightarrow \left( x+\dfrac {2} {x^2} \right)^2 - 6 \left( x+\dfrac {2} {x} \right) + 8 = 0$

$\Rightarrow \left( x+\dfrac {2} {x^2} \right) = \dfrac {6\pm \sqrt{36-32}}{2} = \dfrac {6\pm2} {2} = 4$ or $2$

$\Rightarrow \begin{cases} x+\dfrac {2} {x^2} = 4 & \Rightarrow x^2 - 4x + 2 = 0 &\Rightarrow x = 2\pm \sqrt{2}\\ x+\dfrac {2} {x^2} = 4 & \Rightarrow x^2 - 2x + 2 = 0 &\Rightarrow \text{No real roots} \end {cases}$

Therefore the absolute difference of real roots

$= |2+\sqrt{2}-2+\sqrt{2}| = 2\sqrt{2}= \sqrt{8}\quad \Rightarrow n = \boxed{8}$

$let\quad (x^2+2)^2\quad be\quad a$ $a^2+8x^2 =6xa\longrightarrow a=\dfrac{6x\pm\sqrt{36x^2-32x^2}}{2}\rightarrow a=4x,2x$ $a=4x\longrightarrow x^2-4x+2=0\longrightarrow x=2\pm\sqrt{2}$ $a=2x\longrightarrow x^2-2x+2=0\longrightarrow x=1\pm i$ the real roots $\left|2+\sqrt{2}-(2-\sqrt{2}\right|=\left|2\sqrt{2}\right|=\boxed{\sqrt{8}}$