Simplify the expression (4x8y⁻⁶z)/(16x4y⁻⁴z⁻³) raised to the power of 1/2.

We use the exponent rule $\frac{a^m}{a^n} = a^{m-n}$.

For the numerical coefficients:

$$\frac{4}{16} = \frac{1}{4}$$

For the $x$ terms:

$$\frac{x^8}{x^4} = x^{8-4} = x^4$$

For the $y$ terms:

$$\frac{y^{-6}}{y^{-4}} = y^{-6 - (-4)} = y^{-6+4} = y^{-2}$$

For the $z$ terms (remember $z = z^1$):

$$\frac{z^1}{z^{-3}} = z^{1 - (-3)} = z^{1+3} = z^4$$

Combine these simplified terms:

$$\left(\frac{1}{4} x^4 y^{-2} z^4\right)^{\frac{1}{2}}$$

We use the exponent rule $(ab)^n = a^n b^n$ and $(a^m)^n = a^{mn}$.

$$\left(\frac{1}{4}\right)^{\frac{1}{2}} \left(x^4\right)^{\frac{1}{2}} \left(y^{-2}\right)^{\frac{1}{2}} \left(z^4\right)^{\frac{1}{2}}$$

$$\left(\frac{1}{4}\right)^{\frac{1}{2}} = \sqrt{\frac{1}{4}} = \frac{\sqrt{1}}{\sqrt{4}} = \frac{1}{2}$$ $$\left(x^4\right)^{\frac{1}{2}} = x^{4 \times \frac{1}{2}} = x^2$$ $$\left(y^{-2}\right)^{\frac{1}{2}} = y^{-2 \times \frac{1}{2}} = y^{-1}$$ $$\left(z^4\right)^{\frac{1}{2}} = z^{4 \times \frac{1}{2}} = z^2$$

$$\frac{1}{2} x^2 y^{-1} z^2$$

Since $y^{-1} = \frac{1}{y}$, the expression becomes:

$$\frac{1}{2} x^2 \frac{1}{y} z^2 = \frac{x^2 z^2}{2y}$$

The simplified expression is:

$$\frac{x^2 z^2}{2y}$$