<s>[INST] How many distinct real roots does $ (x^2 + x – 2)^3 + (8–2x^2 )^3 = (x^2 + 3x + 2)^3 $have? 
How many distinct real roots does $ (x^2 + x – 2)^3 + (8–2x^2 )^3 =
  (x^2 + 3x + 2)^3 $have?

If I didn't make any mistake that equation could be reduced to the form:$$(2+6x^2)(x+2)^3=(8–2x^2 )^3$$
However,I don't know how to approach next as more simplification and then forming up the factors again would be more time consuming,I suppose this problem could be solved in a more easy way,any ideas?
 [/INST] 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.
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