2016/106) solve for x $(x^2+2)^2 + 8x^2 = 6x(x^2+2)$

taking the term on RHS to LHS
 $(x^2+2)(x^2 - 6x + 2) + 8(x^2)  = 0$
$(x^2-3x+2 + 3x )(x^2 - 3x + 2 - 3x) + 8x^2 = 0$
or $(x^2-3x+2)^2  - 9x^2 + 8x^2 = 0$
or $(x^2+ 3x +2)^2 - x^2 = 0$
or $(x^2+2x+ 2)(x^2+4x+2) = 0$
$x^2+2x+2=0$ gives $x= -1\pm i$ and $x^2+4x+2=0$ give $2\pm \sqrt{2}$