<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] We are trying to solve the equation
$$(x^2 + x – 2)^3 + (8–2x^2 )^3 -(x^2 + 3x + 2)^3=0.$$
Let $P(x)$ be the polynomial on the left-hand side. Note that $x^2+x-2=(x+2)(x-1)$, and $8-2x^2=-2(x-2)(x+2)$, and $x^2+3x+2=(x+2)(x+1).$
Thus 
$$P(x)=(x+2)^3[(x-1)^3-8(x-2)^3-(x+1)^3]=(x-2)^3Q(x).$$
Expand the remaining cubes. We get
$$Q(x)=-8x^3+42x^2-96x+62.$$
We get very lucky: $x=1$ is a root of $Q(x)$. Divide.
$$Q(x)=(x-1)(-8x^2+34x-62).$$
It is easy to verify, say by the Quadratic Formula, that $8x^2-34x+62=0$ has no real roots.
So there are altogether two distinct real roots, $x=-2$ and $x=1$.
Your calculation: Things would have worked out fine, if after the (correct) simplification of the difference of cubes you had used the fact that $8-2x^2=-2(x+2)(x-2)$. But except in special circumstances, of which this happens to be one, it is easier to deal with a polynomial equation of shape $P(x)=0$ than with one of shape $P_1(x)=P_2(x)$.  
Comment: The question has a quite artificial character, because of the occurrence of an obvious root $x=-2$ of multiplicity at least $3$. Given an "arbitrary" degree $6$ polynomial, the techniques for finding the number of real roots would be quite different. Nowadays, an almost automatic first step is to see what graphing software, or a tool like Wolfram Alpha,  has to say.
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