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The given equation to solve for \( x \) is:

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

with the condition that \( a \neq 0, -1, 1 \).

Let's proceed by simplifying the given equation to solve for \( x \). I will break it down step by step.

First, observe that the left-hand side can be simplified as a difference of cubes:

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

Now, let's equate this to the right-hand side and simplify:

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

I will now simplify further and solve for \( x \).The solution to the given equation is:

\[
x = a^2 + 1
\]

This satisfies the equation under the given condition \( a \neq 0, -1, 1 \).

Would you like more details on the steps of the solution, or do you have any further questions?

Here are five related questions for further exploration:
1. How would the solution change if different restrictions were placed on \( a \)?
2. What happens if \( a = 1 \) or \( a = -1 \), and how does this affect the equation?
3. Can this method be applied to similar equations involving cubic terms?
4. How would the solution differ if the original equation involved \( a^4 \) instead of \( a^3 \)?
5. How does the symmetry of the equation help in simplifying it?

**Tip:** When solving rational equations, always look out for opportunities to factor expressions, as this often leads to easier simplifications.

Solve for x: \[\frac{a^3 - 1}{a^3 + 1} = \frac{a(x - 1) + a^2 - x}{a(x - 1) - a^2 + x}\] with the condition \(a \neq 0, -1, 1\).

This article covers solving a rational equation involving cubic terms. Using the difference of cubes and rational equation techniques, we simplify and solve the equation for x. Suitable for high school or pre-university students.

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