To solve the equation

$\sqrt{\sqrt{(25^0)^3}} \div \frac{3}{4} \div \frac{7}{256}x = \sqrt{\sqrt{\frac{4x^2 + 256}{\sqrt{\left(\frac{120}{5!}\right)^{256}}}}}$

we will follow these steps:

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

1. Simplify $25^0$: Any number raised to the power of 0 is 1, so $25^0 = 1$.
2. Raise 1 to the power of 3: $(25^0)^3 = 1^3 = 1$.
3. Simplify the square root: $\sqrt{\sqrt{1}} = \sqrt{1} = 1$.
4. Divide by the fractions:
   • $\frac{3}{4} \div \frac{7}{256} = \frac{3}{4} \times \frac{256}{7} = \frac{3 \times 256}{4 \times 7} = \frac{768}{28} = \frac{192}{7}$.
   • The LHS is now $x \times \frac{192}{7}$.

So, the LHS simplifies to:

$\frac{192x}{7}$

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

1. Calculate $5!$: $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.
2. Simplify the term inside the square root: Since $\frac{120}{5!} = \frac{120}{120} = 1$.
3. Raise 1 to the power of 256: $1^{256} = 1$.
4. The square root becomes: $\sqrt{1} = 1$.

So, the RHS simplifies to:

$\sqrt{\sqrt{4x^2 + 256}} = \sqrt{\sqrt{4x^2 + 256}}$

Step 3: Equating the Simplified LHS and RHS

$\frac{192x}{7} = \sqrt{\sqrt{4x^2 + 256}}$

Step 4: Solve for $x$

1. Square both sides to remove the square roots:

$\left(\frac{192x}{7}\right)^2 = \sqrt{4x^2 + 256}$

2. Square both sides again:

$\left(\frac{192x}{7}\right)^4 = 4x^2 + 256$

This results in a polynomial equation that can be solved using algebraic methods.

Would you like to continue solving this equation in detail?

Here are five questions related to the concepts we used:

1. What are the properties of exponents that allow us to simplify $25^0$?
2. How do you simplify complex fractions like $\frac{3}{4} \div \frac{7}{256}$?
3. Why does any number raised to the power of zero equal one?
4. How does the factorial function $n!$ affect large numbers?
5. What are the properties of square roots and nested square roots?

Tip: Always simplify expressions inside the parentheses or under the square root before dealing with higher-order operations like powers or products.