1. **State the problem:** We are given the polynomial $$40\sqrt{5} x^{3} - 3\sqrt{3} y^{3}$$ and it is factored as $$(2\sqrt{5} x - \sqrt{3} y)(Ax^{2} + Bxy + Cy^{2}).$$ We need to find the value of $B$. 2. **Recall the formula for factoring:** When factoring a cubic polynomial of the form $a^{3} - b^{3}$, it factors as: $$a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2}).$$ 3. **Identify $a$ and $b$:** Here, $a = 2\sqrt{5} x$ and $b = \sqrt{3} y$. 4. **Write the expanded form:** $$(2\sqrt{5} x - \sqrt{3} y)(A x^{2} + B x y + C y^{2}) = 40\sqrt{5} x^{3} - 3\sqrt{3} y^{3}.$$ Expanding the left side: $$= (2\sqrt{5} x)(A x^{2}) + (2\sqrt{5} x)(B x y) + (2\sqrt{5} x)(C y^{2}) - (\sqrt{3} y)(A x^{2}) - (\sqrt{3} y)(B x y) - (\sqrt{3} y)(C y^{2}).$$ Simplify each term: $$= 2A \sqrt{5} x^{3} + 2B \sqrt{5} x^{2} y + 2C \sqrt{5} x y^{2} - A \sqrt{3} x^{2} y - B \sqrt{3} x y^{2} - C \sqrt{3} y^{3}.$$ 5. **Match coefficients with the original polynomial:** The original polynomial has only $x^{3}$ and $y^{3}$ terms, so the coefficients of $x^{2} y$ and $x y^{2}$ must be zero: - Coefficient of $x^{3}$: $2A \sqrt{5} = 40 \sqrt{5} \implies 2A = 40 \implies A = 20$ - Coefficient of $y^{3}$: $-C \sqrt{3} = -3 \sqrt{3} \implies -C = -3 \implies C = 3$ - Coefficient of $x^{2} y$: $2B \sqrt{5} - A \sqrt{3} = 0$ - Coefficient of $x y^{2}$: $2C \sqrt{5} - B \sqrt{3} = 0$ 6. **Solve for $B$ using the two equations:** From $x^{2} y$ term: $$2B \sqrt{5} = A \sqrt{3} = 20 \sqrt{3}$$ $$B = \frac{20 \sqrt{3}}{2 \sqrt{5}} = 10 \frac{\sqrt{3}}{\sqrt{5}} = 10 \sqrt{\frac{3}{5}}.$$ From $x y^{2}$ term: $$2C \sqrt{5} = B \sqrt{3}$$ $$2 \times 3 \sqrt{5} = B \sqrt{3}$$ $$6 \sqrt{5} = B \sqrt{3}$$ $$B = \frac{6 \sqrt{5}}{\sqrt{3}} = 6 \sqrt{\frac{5}{3}}.$$ 7. **Check consistency:** We have two expressions for $B$: $$10 \sqrt{\frac{3}{5}} \quad \text{and} \quad 6 \sqrt{\frac{5}{3}}.$$ Simplify both: $$10 \sqrt{\frac{3}{5}} = 10 \frac{\sqrt{3}}{\sqrt{5}} = \frac{10 \sqrt{3}}{\sqrt{5}}$$ $$6 \sqrt{\frac{5}{3}} = 6 \frac{\sqrt{5}}{\sqrt{3}} = \frac{6 \sqrt{5}}{\sqrt{3}}$$ Multiply numerator and denominator to rationalize: $$10 \frac{\sqrt{3}}{\sqrt{5}} = 10 \frac{\sqrt{3} \sqrt{5}}{5} = 10 \frac{\sqrt{15}}{5} = 2 \sqrt{15}$$ $$6 \frac{\sqrt{5}}{\sqrt{3}} = 6 \frac{\sqrt{5} \sqrt{3}}{3} = 6 \frac{\sqrt{15}}{3} = 2 \sqrt{15}$$ Both equal $2 \sqrt{15}$, so $B = 2 \sqrt{15}$. **Final answer:** $$\boxed{2 \sqrt{15}}$$