The expression given is:

$((x^4y^0)^{-4}) \cdot 2y^0$

Let’s break it down step-by-step.

Step 1: Simplify $y^0$

Any non-zero term raised to the power of zero is 1, so $y^0 = 1$. Therefore, $x^4y^0$ simplifies to $x^4$.

Step 2: Apply the Power of -4

Now we have:

$(x^4)^{-4} \cdot 2y^0$

Since $y^0 = 1$, we can ignore it, and it simplifies further to:

$(x^4)^{-4} \cdot 2$

Step 3: Apply the Exponent

When a power is raised to another power, we multiply the exponents:

$(x^4)^{-4} = x^{4 \cdot (-4)} = x^{-16}$

So the expression now becomes:

$x^{-16} \cdot 2$

Step 4: Write the Answer

This is simplified as:

$2x^{-16}$

or, if you want to express it with a positive exponent,

$\frac{2}{x^{16}}$

Final Answer

$2x^{-16} \quad \text{or} \quad \frac{2}{x^{16}}$

Would you like more details on any step?

Here are five questions related to this concept:

1. What happens if we change the exponent from $-4$ to a positive $4$?
2. How would the solution change if $y$ were non-zero?
3. What would the expression look like if we had $(x^4y^2)^{-4} \cdot 2y^2$?
4. How does exponentiation affect other types of mathematical operations?
5. Why is $y^0 = 1$ for any non-zero value of $y$?

Tip: Remember that any non-zero term raised to the zero power is always 1.