Let's revisit the problem and find the correct solution.

Given the equation:

\[ X^2 - \frac{1}{X} + 2 = 0 \]

First, we need to find \( X \).

Multiply through by \( X \) to clear the fraction:

\[ X^3 - 1 + 2X = 0 \]

\[ X^3 + 2X - 1 = 0 \]

Solving this cubic equation exactly can be challenging. However, we can try to find a pattern or a method to directly determine \( X^2 + \frac{1}{X^2} \).

We start by multiplying the original equation by \( X \):

\[ X^3 + 2X - 1 = 0 \]

Now, notice that we can try a different approach by squaring \( X - \frac{1}{X} \):

\[ \left(X - \frac{1}{X}\right)^2 = X^2 - 2 + \frac{1}{X^2} \]

From the original equation, isolate \( \frac{1}{X} \):

\[ X^2 + 2 = \frac{1}{X} \]

So:

\[ X - \left(X^2 + 2\right) = 0 \]

This means \( X - \frac{1}{X} = 0 \):

This means \(X = 1/X\).

Let's set \( y = X - \frac{1}{X} \):

Now:

\[ y = X - X^2 - 2 \]

We know that \( X \cdot \frac{1}{X} = 1 \), thus

\[ \left(X - \frac{1}{X}\right)^2 + 2 \]

Substituting our given values back, we realize:

\[ X^2 + \frac{1}{X^2} = y^2 + 2 \]

However, we must solve \(X\) to find the exact value. 

Trying for simpler roots:

The roots, therefore, must simplify properly to \(1\).

Thus, \( X = 1 \) or \( X = -1\).

Thus checking back:

\[ X^2 - \frac{1}{X} + 2 = 0 \]

\( X = 1\) or

\[ -\frac{1}{X} + 2 = 0 \].

Thus:

\[ X = -2\]

\[\Rightarrow  X^2 + \frac{1}{X^2}\].

Thus:

This means:

Answer is:

Thus we get finally: 2

**Would you like more details or have any questions?**

Here are five questions you can ask next:
1. How to solve cubic equations?
2. Can you explain the steps to simplify a given polynomial?
3. What are the properties of quadratic equations?
4. How can we factorize complex algebraic expressions?
5. What are the methods to solve higher degree polynomials?

**Tip:** When solving polynomial equations, always check for possible factorizations or simplifications that can make the solving process easier.

Learn how to solve the cubic equation X^3 + 2X - 1 = 0 step-by-step. This problem involves finding roots of cubic equations and applying cubic equation solving methods. Suitable for advanced high school students (Grades 11-12).

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