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The equation provided is:

\[
\frac{1}{x^2} - 4 \cdot \frac{1}{x^4} = 0
\]

To solve this equation, let's follow these steps:

1. Rewrite \(\frac{1}{x^2}\) as a variable to simplify. Let \(y = \frac{1}{x^2}\), then \(\frac{1}{x^4} = y^2\).

   Substituting these, the equation becomes:
   \[
   y - 4y^2 = 0
   \]

2. Factor out \(y\):
   \[
   y(1 - 4y) = 0
   \]

3. Set each factor equal to zero:
   - \(y = 0\)
   - \(1 - 4y = 0\)

   Solving these:
   - For \(y = 0\): \(y = \frac{1}{x^2} = 0\), which has no solution since \(x\) cannot be infinite.
   - For \(1 - 4y = 0\): \(y = \frac{1}{4}\)

4. Substitute back \(y = \frac{1}{x^2}\) and set it equal to \(\frac{1}{4}\):
   \[
   \frac{1}{x^2} = \frac{1}{4}
   \]

5. Solve for \(x\):
   \[
   x^2 = 4
   \]
   \[
   x = \pm 2
   \]

### Solution:
The solutions to the equation are:
\[
x = 2 \quad \text{and} \quad x = -2
\]

Would you like more details on this solution, or have any questions? 

Here are five related questions that may deepen understanding:
1. What happens if there is an additional constant term in the equation?
2. How would the approach change if the equation had \(\frac{1}{x^3}\) instead of \(\frac{1}{x^4}\)?
3. Can similar techniques solve equations involving higher powers of \(x\) in the denominator?
4. How can substitution methods be generalized for more complex equations?
5. What role does factoring play in solving equations like this?

**Tip:** In equations with fractions, substitution can simplify terms and make factoring easier.

Solve \(\frac{1}{x^2} - 4 \cdot \frac{1}{x^4} = 0\)

Explore a step-by-step solution to the equation \(\frac{1}{x^2} - 4 \cdot \frac{1}{x^4} = 0\). This solution involves substitution and factoring techniques, focusing on algebraic concepts suitable for students in Grades 10-12.

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