$\left ( x^{2} + 2 \right )\sqrt{x^{2} + 1} = 8x^{3} + 2x$

$\Rightarrow \left ( x^{2} + 2 \right )^{2}.\left ( x^{2} + 1 \right ) = 64x^{6} + 32x^{4} + 4x^{2}$

$\Leftrightarrow \left ( x^{4} + 4x^{2} + 4 \right ).\left ( x^{2} + 1 \right ) = 64x^{6} + 32x^{4} + 4x^{2}$

$\Leftrightarrow x^{6} + x^{4} + 4x^{4} + 4x^{2} + 4x^{2} + 4 = 64x^{6} + 32x^{4} + 4x^{2}$

$\Leftrightarrow x^{6} + x^{4} + 4x^{4} + 4x^{2} + 4 = 64x^{6} + 32x^{4}$

$\Leftrightarrow x^{6} + 5x^{4} + 4x^{2} + 4 = 64x^{6} + 32x^{4}$

$\Leftrightarrow x^{6} + 5x^{4} + 4x^{2} + 4 - 64x^{6} - 32x^{4} = 0$

$\Leftrightarrow -63x^{6} + 21x^{4} - 48x^{4} + 16x^{2} - 12x^{2} + 4 = 0$

$\Leftrightarrow -21x^{4}.\left ( 3x^{2} - 1 \right ) - 16x^{2}.\left ( 3x^{2} - 1 \right ) - 4.\left ( 3x^{2} - 1 \right ) = 0$

$\Leftrightarrow -\left ( 3x^{2} - 1 \right ).\left ( 21x^{4} + 16x^{2} + 4 \right ) = 0$

$\Leftrightarrow \left ( 3x^{2} - 1 \right ).\left ( 21x^{4} + 16x^{2} + 4 \right ) = 0$

$\Leftrightarrow \left[ \begin{array}{l}3x^{2} - 1 = 0\\21x^{4} + 16x^{2} + 4 = 0\end{array} \right.$ 

$\Leftrightarrow x = \dfrac{\sqrt{3}}{3}$