To find the cube root of $-27x^9y^5$, we take the cube root of each factor separately:

1. Cube root of the constant term: $\sqrt[3]{-27} = -3$

2. Cube root of the $x$ term: $\sqrt[3]{x^9} = x^{9/3} = x^3$

3. Cube root of the $y$ term: $\sqrt[3]{y^5} = y^{5/3}$

Combining these, the cube root of $-27x^9y^5$ is: $-3x^3y^{5/3}$

Final Answer:

$\sqrt[3]{-27x^9y^5} = -3x^3y^{5/3}$

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Here are some related questions to deepen understanding:

1. What is the cube root of a negative variable expression, such as $-64a^{12}b^6$?
2. How does taking an even root (e.g., square root) differ when dealing with negative numbers compared to odd roots?
3. What would be the cube root of $27x^{12}y^9$?
4. How would the cube root of a fraction, like $\frac{-27x^9}{y^6}$, be calculated?
5. How does exponent notation change when roots are applied to variables?

Tip: For roots of variable expressions, remember to apply the root to each term individually for clarity and accuracy.