To find the monomial whose cube is given by the expression \(-0.001a^{3n + 3}\), we can set up the equation as follows:
Let the monomial be \(x\).
Then we have:
\[ x^3 = -0.001a^{3n + 3} \]
To find \(x\), we take the cube root of both sides:
\[ x = \sqrt[3]{-0.001} \cdot \sqrt[3]{a^{3n + 3}} \]
Calculating the cube root of \(-0.001\):
\[ \sqrt[3]{-0.001} = -0.1 \]
Now, for the second part:
\[ \sqrt[3]{a^{3n + 3}} = a^{(3n + 3)/3} = a^{n + 1} \]
Putting it all together, we find the monomial:
\[ x = -0.1 a^{n + 1} \]
Thus, the monomial is:
\[ -0.1 a^{n + 1} \]