Solution of the equation${\left(x^2+2\right)}^2+8x^2=6x\left(x^2+2\right)$

The given equation reduces to

 $\begin{array}{l}{a}^2+8x^2-6ax=0\text{ where }a=\left(x^2+2\right)\\ \Rightarrow 8x^2-6ax+a^2=0\\ \Rightarrow 8x^2-4ax-2ax+a^2=0\\ \Rightarrow 4x(2x-a)-a(2x-a)=0\\ \Rightarrow (4x-a)(2x-a)=0\\ \therefore a=2x,4x\end{array}$

When $a=2x,$ then $x^2+2=2x$

$\begin{array}{l}\Rightarrow x^2-2x+2=0\\ \Rightarrow x=\frac{2\pm \sqrt{4-8}}{2}=1\pm i\end{array}$

When $a=4x,$ then $x^2+2=4x$

$\begin{array}{l}\Rightarrow x^2-4x+2=0\\ \Rightarrow x=\frac{4\pm \sqrt{16-8}}{2}=2\pm \sqrt{2}\end{array}$

Hence the solutions are $\left\{\begin{array}{l}1\pm i, 2\pm \sqrt{2}\end{array}\right\}$