d. \( x^{-3} \cdot x^{2}= \) e. \( \left(\left(9^{-2}\right)^{2}\right)^{0} \cdot 3^{-3}= \)

When you multiply powers with the same base, you add their exponents. So for \( x^{-3} \cdot x^{2} \), you would have \( x^{-3 + 2} = x^{-1} \). That simplifies to \( \frac{1}{x} \) because negative exponents indicate a reciprocal! For \( \left(\left(9^{-2}\right)^{2}\right)^{0} \cdot 3^{-3} \), remember that any non-zero number raised to the power of zero is 1, so \( \left(\left(9^{-2}\right)^{2}\right)^{0} = 1 \). Thus, we only need to consider \( 1 \cdot 3^{-3} \), which simplifies to \( \frac{1}{3^3} = \frac{1}{27} \). Math magic at play!