1. **State the problem:** Expand and simplify the expression $$(x - 4y)^5$$ using the binomial theorem. 2. **Recall the binomial theorem formula:** $$ (a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k $$ where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ is the binomial coefficient. 3. **Apply the formula to our problem:** Here, $a = x$, $b = -4y$, and $n = 5$. 4. **Write the expansion:** $$ (x - 4y)^5 = \sum_{k=0}^5 \binom{5}{k} x^{5-k} (-4y)^k $$ 5. **Calculate each term:** - For $k=0$: $\binom{5}{0} x^5 (-4y)^0 = 1 \cdot x^5 \cdot 1 = x^5$ - For $k=1$: $\binom{5}{1} x^4 (-4y)^1 = 5 \cdot x^4 \cdot (-4y) = -20x^4y$ - For $k=2$: $\binom{5}{2} x^3 (-4y)^2 = 10 \cdot x^3 \cdot 16y^2 = 160x^3y^2$ - For $k=3$: $\binom{5}{3} x^2 (-4y)^3 = 10 \cdot x^2 \cdot (-64y^3) = -640x^2y^3$ - For $k=4$: $\binom{5}{4} x^1 (-4y)^4 = 5 \cdot x \cdot 256y^4 = 1280xy^4$ - For $k=5$: $\binom{5}{5} x^0 (-4y)^5 = 1 \cdot 1 \cdot (-1024y^5) = -1024y^5$ 6. **Combine all terms:** $$ x^5 - 20x^4y + 160x^3y^2 - 640x^2y^3 + 1280xy^4 - 1024y^5 $$ 7. **Final answer:** $$ \boxed{x^5 - 20x^4y + 160x^3y^2 - 640x^2y^3 + 1280xy^4 - 1024y^5} $$