Don't expand

Let $y = x^2+1$ , then:

$\begin{aligned} (x^2+x+1)^3-(x^2+1)^3-x^3 & = (y+x)^3 - y^3 - x^3 \\ & = y^3 + 3y^2x +3yx^2 + x^3 - y^3 - x^3 \\ & = 3xy(x+y) \\ & = 3x(x^2+1)(x^2+x+1) \end{aligned}$

Similarly,

$\begin{aligned} (x^2-x+1)^3-(x^2+1)^3+x^3 & = (y-x)^3 - y^3 + x^3 \\ & = y^3 - 3y^2x + 3yx^2 - x^3 - y^3 + x^3 \\ & = 3xy(x-y) \\ & = -3x(x^2+1)(x^2-x+1) \end{aligned}$

And,

$\begin{aligned} (x^4+x^2+1)^3-(x^2+1)^3-x^3 & = 3x^2(x^4+1)(x^4+x^2+1) \end{aligned}$

Therefore,

$\begin{aligned} LHS & = [3xy(x+y)] [3xy(x-y)] = 9x^2y^2(x^2-y^2) = - 9x^2(x^2+1)^2(x^4+x^2+1) \end{aligned}$

$\begin{aligned} \Rightarrow - 9x^2(x^2+1)^2(x^4+x^2+1) & = 9x^2(x^4+1)(x^4+x^2+1) \\ x^2(x^4+x^2+1)(x^4 + 1 +x^4+2x^2+1) & = 0 \\ x^2(x^4+x^2+1)^2 & = 0 \\ \Rightarrow x & = 0 \quad \quad \small \color{#3D99F6}{x^4+x^2+1 \text{ has no real root.}} \end{aligned}$

Therefore, the sum of all integral values of $x$ is $\boxed{0}$ .

To avoid expanding the expression, we can choose appropriate substitutions and use the identity $(a+b)^3-a^3-b^3 = 3ab(a+b)$ .

On the LHS, let $m = x^2+1$ and on the RHS, let $n = x^4+1$ .

$[(m+x)^3-m^3-x^3][(m-x)^3-m^3+x^3] = 3[(n+x^2)^3-n^3-(x^2)^3]$ $[3mx(x+m)]*[3mx(x-m)] = 3[3nx^2(n+x^2)]$

Dividing both sides by 9, $m^2x^2(x^2-m^2) = nx^2(n+x^2)$

which gives us x = 0. Next we note that $n = x^4+1 = (x^2+1)^2-2x^2 = m^2-2x^2$ . Substituting this back into the equation and dividing by x^2,

$m^2(x^2-m^2) = (m^2-2x^2)(m^2-x^2)$ $(x^2-m^2)(2m^2-2x^2) = 0$ $-2(x-m)^2(x+m)^2=0$ $x=m$ or $x=-m$ gives us $x^2-x+1 = 0$ or $x^2+x+1 = 0$ which have no integer solutions.

Hence the only integer solution is $x=0$ and the sum of integral values of x is $\boxed {0}$ .