Question

\\(2y^3)^2 \cdot 2yx^{-1} \cdot 2y^{-1}\\

Explanation:

Expand the squared term

$$ (2y^3)^2 = 2^2 \cdot (y^3)^2 = 4y^6 $$

Multiply the remaining terms

$$ 4y^6 \cdot 2yx^{-1} \cdot 2y^{-1} = (4 \cdot 2 \cdot 2) \cdot x^{-1} \cdot (y^6 \cdot y^1 \cdot y^{-1}) $$

Simplify exponents and write with positive exponents

$$ 16 \cdot x^{-1} \cdot y^{6+1-1} = 16x^{-1}y^6 = \frac{16y^6}{x} $$

Answer:

$$\frac{16y^6}{x}$$