To simplify $\sqrt[3]{40x^7y^8}$, let’s break down each part:

1. Factor out constants and simplify them if possible. The number $40$ can be factored into $8 \times 5$, where $8$ is a perfect cube: [ \sqrt[3]{40} = \sqrt[3]{8 \times 5} = 2 \cdot \sqrt[3]{5} ]

2. Simplify the variable parts under the cube root:

   • For $x^7$: $\sqrt[3]{x^7} = x^2 \cdot \sqrt[3]{x}$, since $x^6$ (a multiple of 3) becomes $x^2$, and $x$ is left under the root.
   • For $y^8$: $\sqrt[3]{y^8} = y^2 \cdot \sqrt[3]{y^2}$, since $y^6$ (a multiple of 3) becomes $y^2$, and $y^2$ remains under the root.

Putting it all together: [ \sqrt[3]{40x^7y^8} = 2x^2y^2 \cdot \sqrt[3]{5xy^2} ]

So, the simplified form is: $2x^2y^2 \cdot \sqrt[3]{5xy^2}$

Would you like more details, or do you have any questions?

1. How would the process differ if it were a square root instead of a cube root?
2. Can cube roots of variables always be simplified in a similar way?
3. What happens if the exponents were lower than the root index (e.g., square root of $x^2$)?
4. How does factoring out constants help in simplifying radicals?
5. Are there alternative methods to simplify cube roots?

Tip: Always look for perfect powers (like $8$ in this example) when simplifying roots.