To solve the equation

\[
\frac{x + 1}{x^2 + x + 1} - \frac{x - 1}{x^2 - x + 1} = \frac{2(x + 2)^2}{x^6 - 1},
\]

we will start by simplifying each term step by step.

### Step 1: Simplify the Left-Hand Side (LHS)

First, let's find a common denominator for the fractions on the LHS:

\[
\frac{x + 1}{x^2 + x + 1} - \frac{x - 1}{x^2 - x + 1}.
\]

The common denominator is \((x^2 + x + 1)(x^2 - x + 1)\).

Rewrite each fraction with the common denominator:

\[
\frac{(x + 1)(x^2 - x + 1) - (x - 1)(x^2 + x + 1)}{(x^2 + x + 1)(x^2 - x + 1)}.
\]

Expand the numerators:

\[
(x + 1)(x^2 - x + 1) = x^3 - x^2 + x + x^2 - x + 1 = x^3 + 1,
\]

\[
(x - 1)(x^2 + x + 1) = x^3 + x^2 + x - x^2 - x - 1 = x^3 - 1.
\]

So the LHS becomes:

\[
\frac{x^3 + 1 - (x^3 - 1)}{(x^2 + x + 1)(x^2 - x + 1)} = \frac{x^3 + 1 - x^3 + 1}{(x^2 + x + 1)(x^2 - x + 1)} = \frac{2}{(x^2 + x + 1)(x^2 - x + 1)}.
\]

### Step 2: Simplify the Right-Hand Side (RHS)

The RHS is:

\[
\frac{2(x + 2)^2}{x^6 - 1}.
\]

Notice that \(x^6 - 1\) can be factored as:

\[
x^6 - 1 = (x^3 - 1)(x^3 + 1).
\]

Thus, the RHS becomes:

\[
\frac{2(x + 2)^2}{(x^3 - 1)(x^3 + 1)}.
\]

### Step 3: Compare LHS and RHS

We need to compare:

\[
\frac{2}{(x^2 + x + 1)(x^2 - x + 1)}
\]

with

\[
\frac{2(x + 2)^2}{(x^3 - 1)(x^3 + 1)}.
\]

Recall that:

\[
x^3 - 1 = (x - 1)(x^2 + x + 1)
\]

and

\[
x^3 + 1 = (x + 1)(x^2 - x + 1).
\]

Thus,

\[
(x^3 - 1)(x^3 + 1) = (x - 1)(x^2 + x + 1)(x + 1)(x^2 - x + 1).
\]

So,

\[
\frac{2}{(x^2 + x + 1)(x^2 - x + 1)} = \frac{2}{(x^2 + x + 1)(x^2 - x + 1)}.
\]

This implies that:

\[
(x + 2)^2 = 1.
\]

Solving for \(x\):

\[
x + 2 = \pm 1.
\]

Thus,

\[
x + 2 = 1 \implies x = -1,
\]

or

\[
x + 2 = -1 \implies x = -3.
\]

### Step 4: Verify Solutions

We need to check if \(x = -1\) and \(x = -3\) satisfy the original equation.

#### For \(x = -1\):

LHS:

\[
\frac{-1 + 1}{(-1)^2 - 1 + 1} - \frac{-1 - 1}{(-1)^2 + 1 + 1} = \frac{0}{1} - \frac{-2}{3} = 0 + \frac{2}{3} = \frac{2}{3}.
\]

RHS:

\[
\frac{2(-1 + 2)^2}{(-1)^6 - 1} = \frac{2(1)^2}{1 - 1} = \frac{2}{0},
\]

which is undefined. So, \(x = -1\) is not a valid solution.

#### For \(x = -3\):

LHS:

\[
\frac{-3 + 1}{(-3)^2 - 3 + 1} - \frac{-3 - 1}{(-3)^2 + 3 + 1} = \frac{-2}{9 - 3 + 1} - \frac{-4}{9 + 3 + 1} = \frac{-2}{7} + \frac{4}{13}.
\]

RHS:

\[
\frac{2(-3 + 2)^2}{(-3)^6 - 1} = \frac{2(-1)^2}{729 - 1} = \frac{2}{728} = \frac{1}{364}.
\]

Since \(\frac{-2}{7} + \frac{4}{13} \neq \frac{1}{364}\), \(x = -3\) is not a valid solution either.

Thus, the equation has no valid solutions.