To find the monomial \( x \) such that
\[ x^3 = -\frac{27}{64} a^9 \cdot a^6, \]
we first simplify the right side of the equation.
Using the property of exponents that states \( a^m \cdot a^n = a^{m+n} \), we combine the exponents on \( a \):
\[ a^9 \cdot a^6 = a^{9+6} = a^{15}. \]
So, our expression simplifies to:
\[ x^3 = -\frac{27}{64} a^{15}. \]
Next, we take the cube root of both sides to find \( x \):
\[ x = \sqrt[3]{-\frac{27}{64}} \cdot \sqrt[3]{a^{15}}. \]
Now, we can simplify each term separately:
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To simplify \( \sqrt[3]{-\frac{27}{64}} \):
- The cube root of -27 is -3 (as \((-3)^3 = -27\)).
- The cube root of 64 is 4 (as \(4^3 = 64\)).
- Thus, \( \sqrt[3]{-\frac{27}{64}} = -\frac{3}{4} \).
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To simplify \( \sqrt[3]{a^{15}} \):
- Since the cube root of \( a^{15} \) is \( a^{15/3} = a^5 \).
Putting it all together, we have:
\[ x = -\frac{3}{4} a^5. \]
Thus, the monomial \( x \) is
\[ \boxed{-\frac{3}{4} a^5}. \]