Question

\\\frac{2^{-8}x^{-2}y^3}{2^{-4}x^{-5}y^6}\\

Explanation:

Apply the quotient rule for exponents

$$ \frac{2^{-8}x^{-2}y^3}{2^{-4}x^{-5}y^6} = 2^{-8 - (-4)} \cdot x^{-2 - (-5)} \cdot y^{3 - 6} $$

Simplify the exponents

$$ 2^{-8 + 4} \cdot x^{-2 + 5} \cdot y^{3 - 6} = 2^{-4} \cdot x^3 \cdot y^{-3} $$

Convert negative exponents to positive exponents

$$ 2^{-4} \cdot x^3 \cdot y^{-3} = \frac{x^3}{2^4 y^3} = \frac{x^3}{16y^3} $$

Answer:

$$\frac{x^3}{16y^3}$$