Simplify the expression $(\frac{x^3 \cdot 2x^2 y^{-3}}{yx^0})^4$.

1. Problem Description

Simplify the expression $(\frac{x^3 \cdot 2x^2 y^{-3}}{yx^0})^4$.

2. Solution Steps

First, simplify the expression inside the parentheses.

We have $x^0 = 1$.

So the expression becomes $(\frac{x^3 \cdot 2x^2 y^{-3}}{y})^4$.

Multiplying the terms in the numerator yields:

$x^3 \cdot 2x^2 y^{-3} = 2x^{3+2} y^{-3} = 2x^5 y^{-3}$.

Now, the expression inside the parentheses is:

$\frac{2x^5 y^{-3}}{y}$.

Recall that $\frac{y^a}{y^b} = y^{a-b}$. Thus $\frac{y^{-3}}{y^1} = y^{-3-1} = y^{-4}$.

Therefore, $\frac{2x^5 y^{-3}}{y} = 2x^5 y^{-4}$.

Now we can raise this to the power of

4. $(2x^5 y^{-4})^4$.

Recall that $(ab)^n = a^n b^n$. Also $(x^a)^b = x^{a \cdot b}$.

Thus, $(2x^5 y^{-4})^4 = 2^4 (x^5)^4 (y^{-4})^4 = 16 x^{5 \cdot 4} y^{-4 \cdot 4} = 16 x^{20} y^{-16}$.

Also, we can write $y^{-16} = \frac{1}{y^{16}}$.

Therefore, $16x^{20} y^{-16} = \frac{16x^{20}}{y^{16}}$.

3. Final Answer

$\frac{16x^{20}}{y^{16}}$