Using the idea in Aryabhata's comment, since
$$\left( 8-2x^{2}\right) ^{3}=2^{3}\left( 4-x^{2}\right) ^{3}=8
(2-x)^{3}(2+x)^{3},$$
your second equation
$$\left( 2+6x^{2}\right) \left( x+2\right) ^{3}=\left( 8-2x^{2}\right) ^{3}\tag{1}$$
is equivalent to
$$\begin{eqnarray*}
\left( 2+6x^{2}\right) \left( x+2\right) ^{3} &=&8( 2-x)^{3}(2+x)
^{3}.\tag{2}
\end{eqnarray*}$$
So the factor $(x+2)^3$ in both sides yields the real triple root $x=-2$. For $x\ne -2$ this equation $(2)$ is equivalent to
$$\begin{eqnarray*}
\left( 2+6x^{2}\right) &=&-8\left( x-2\right) ^{3}\\
\left( 2+6x^{2}\right) +8\left( x-2\right) ^{3} &=&0 \\
\left( 1+3x^{2}\right) +4\left( x-2\right) ^{3} &=&0\text{.}
\end{eqnarray*}$$
By inspection we see that $x=1$ is a root and by long division or Ruffini's rule, we find:
$$
\begin{eqnarray*}
\left( 1+3x^{2}\right) +4\left( x-2\right) ^{3} =4x^{3}-21x^{2}+48x-31=\left( x-1\right) \left( 4x^{2}-17x+31\right).
\end{eqnarray*}$$
Alternatively we could have applied the rational zero theorem: all the rational roots of the equation $$d_{n}x^{n}+d_{n-1}x^{n-1}+\ldots +d_{0}=0,$$ where all the coefficients are integers, are of the form $\frac{p}{q}=\frac{\text{a factor
of }d_{0}}{\text{a factor of }d_{n}}$.
Hence, for $x\ne -2$, the equation $(1)$ is equivalent to
$$\left( x-1\right) \left( 4x^{2}-17x+31\right) =0.\tag{3}$$
Since the discriminant of the quadratic term is negative $$\Delta
=17^{2}-4\times 4\times 31=-207<0,$$ the original equation has the real
solution $x=1$ due to the factor $(x-1)$ and the triple real solution $x=-2$ due to the common factor $(x+2)^3$. In the total, two distinct real roots.