You have the expression:

$\left( a^{2x} - \sqrt{2} a^{2x} + 1 \right) \left( a^{2x} + \sqrt{2} a^{2x} + 1 \right)$

This is a product of two binomial expressions. We can simplify it by using the distributive property (or FOIL method: First, Outside, Inside, Last). Let's expand it:

Step-by-step Expansion:

1. First terms: Multiply the first terms of both binomials: $a^{2x} \cdot a^{2x} = (a^{2x})^2 = a^{4x}$

2. Outside terms: Multiply the outside terms: $a^{2x} \cdot \sqrt{2} a^{2x} = \sqrt{2} (a^{2x})^2 = \sqrt{2} a^{4x}$

3. Inside terms: Multiply the inside terms: $- \sqrt{2} a^{2x} \cdot a^{2x} = - \sqrt{2} (a^{2x})^2 = - \sqrt{2} a^{4x}$

4. Last terms: Multiply the last terms: $(- \sqrt{2} a^{2x}) \cdot (\sqrt{2} a^{2x}) = -(\sqrt{2})^2 a^{4x} = -2 a^{4x}$ and $1 \cdot 1 = 1$

Combining the terms:

Now add all the terms together: $a^{4x} + \sqrt{2} a^{4x} - \sqrt{2} a^{4x} - 2 a^{4x} + 1$

Notice that $\sqrt{2} a^{4x}$ and $- \sqrt{2} a^{4x}$ cancel each other out, so we are left with:

$a^{4x} - 2 a^{4x} + 1$

This simplifies to:

$- a^{4x} + 1$

Thus, the simplified form of the given expression is:

$1 - a^{4x}$

Would you like more details or have any questions?

Here are some related questions:

1. What is the general process for multiplying binomials?
2. How does the distributive property apply to polynomial expansion?
3. Can this expression be factored further under certain conditions?
4. How does this expansion change if different exponents are used?
5. How does this apply to solving algebraic equations?

Tip: Always check for terms that cancel out during expansions, as it simplifies the final expression significantly.