1. **State the problem:** Find the 2nd term in the expansion of $$\left(\frac{3}{x^5} - 4x^4\right)^4$$ using the binomial theorem. 2. **Recall the binomial theorem formula:** $$\left(a + b\right)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$$ The $k$-th term (starting from $k=0$) is: $$T_{k+1} = \binom{n}{k} a^{n-k} b^k$$ 3. **Identify terms:** Let $a = \frac{3}{x^5} = 3x^{-5}$ and $b = -4x^4$, with $n=4$. 4. **Find the 2nd term:** This corresponds to $k=1$: $$T_2 = \binom{4}{1} (3x^{-5})^{4-1} (-4x^4)^1$$ 5. **Calculate binomial coefficient:** $$\binom{4}{1} = 4$$ 6. **Calculate powers:** $$(3x^{-5})^3 = 3^3 x^{-15} = 27 x^{-15}$$ $$(-4x^4)^1 = -4 x^4$$ 7. **Multiply all parts:** $$T_2 = 4 \times 27 x^{-15} \times (-4 x^4) = 4 \times 27 \times (-4) \times x^{-15+4} = -432 x^{-11}$$ 8. **Final answer:** $$\boxed{-432 x^{-11}}$$ This matches option A.