To solve the problem, we need to simplify the expression $\sqrt[3]{40x^6y^5}$.

Step 1: Break Down the Radicand

The radicand is $40x^6y^5$. Let's break this into smaller parts:

$40x^6y^5 = 8 \times 5 \times x^6 \times y^3 \times y^2$

Here, $8$ is a perfect cube ($8 = 2^3$), and $x^6$ can be written as $(x^2)^3$.

Step 2: Apply the Cube Root

Now apply the cube root to each part:

[ \sqrt[3]{8 \times 5 \times x^6 \times y^5} = \sqrt[3]{8} \times \sqrt[3]{5} \times \sqrt[3]{x^6} \times \sqrt[3]{y^5} ]

Simplify each part:

• $\sqrt[3]{8} = 2$
• $\sqrt[3]{x^6} = x^2$
• $\sqrt[3]{y^5} = y \times \sqrt[3]{y^2}$

So the expression simplifies to:

$2x^2y \sqrt[3]{5y^2}$

Step 3: Match with the Given Options

Looking at the options provided:

• The second option matches exactly: $\boxed{2x^2y \sqrt[3]{5y^2}}$.

Thus, the correct answer is the second option.

Would you like further details or have any other questions?

Here are 5 related questions you might find interesting:

1. How do you simplify expressions under different roots, such as square roots and fourth roots?
2. What are the properties of exponents that help simplify radical expressions?
3. Can you apply the same process to simplify expressions involving higher powers, like $x^9$ or $y^{12}$?
4. How do you deal with negative exponents when simplifying radical expressions?
5. What happens if the radicand is a fraction or involves multiple variables with different exponents?

Tip: Always factor out perfect powers (like squares, cubes) when simplifying radical expressions, as it makes the process smoother.