Question: Simplify.
$\frac{8w^4}{16v^2w^4 - 4wy^3}$
1. **State the problem:** Simplify the expression $$\frac{8w^4}{16v^2w^4 - 4wy^3}$$.
2. **Factor the denominator:**
$$16v^2w^4 - 4wy^3 = 4w(4v^2w^3 - y^3)$$
3. **Rewrite the expression:**
$$\frac{8w^4}{4w(4v^2w^3 - y^3)}$$
4. **Simplify the fraction by canceling common factors:**
$$\frac{\cancel{8}w^{4}}{\cancel{4}w(4v^2w^3 - y^3)} = \frac{2w^{4}}{w(4v^2w^3 - y^3)}$$
5. **Cancel $w$ in numerator and denominator:**
$$\frac{2w^{\cancel{4}}}{\cancel{w}(4v^2w^3 - y^3)} = \frac{2w^{3}}{4v^2w^3 - y^3}$$
6. **Recognize the denominator as a difference of cubes:**
$$4v^2w^3 - y^3 = (\sqrt[3]{4}vw)^3 - y^3$$
7. **Apply difference of cubes formula:**
$$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$
where $$a = \sqrt[3]{4}vw$$ and $$b = y$$.
8. **Factor denominator:**
$$ (\sqrt[3]{4}vw - y)((\sqrt[3]{4}vw)^2 + (\sqrt[3]{4}vw)y + y^2) $$
9. **Final simplified form:**
$$\frac{2w^{3}}{(\sqrt[3]{4}vw - y)((\sqrt[3]{4}vw)^2 + (\sqrt[3]{4}vw)y + y^2)}$$
This is the simplified expression.