1. **State the problem:** Expand the expression $$(2a - 1)^6$$ using the binomial theorem. 2. **Formula used:** The binomial theorem states that $$ (x - y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k} (-y)^k $$ where $\binom{n}{k}$ is the binomial coefficient. 3. **Apply the formula:** Here, $x = 2a$, $y = 1$, and $n = 6$. 4. **Calculate each term:** $$ (2a - 1)^6 = \sum_{k=0}^6 \binom{6}{k} (2a)^{6-k} (-1)^k $$ Calculate terms one by one: - For $k=0$: $\binom{6}{0}(2a)^6(-1)^0 = 1 \times 64a^6 \times 1 = 64a^6$ - For $k=1$: $\binom{6}{1}(2a)^5(-1)^1 = 6 \times 32a^5 \times (-1) = -192a^5$ - For $k=2$: $\binom{6}{2}(2a)^4(-1)^2 = 15 \times 16a^4 \times 1 = 240a^4$ - For $k=3$: $\binom{6}{3}(2a)^3(-1)^3 = 20 \times 8a^3 \times (-1) = -160a^3$ - For $k=4$: $\binom{6}{4}(2a)^2(-1)^4 = 15 \times 4a^2 \times 1 = 60a^2$ - For $k=5$: $\binom{6}{5}(2a)^1(-1)^5 = 6 \times 2a \times (-1) = -12a$ - For $k=6$: $\binom{6}{6}(2a)^0(-1)^6 = 1 \times 1 \times 1 = 1$ 5. **Combine all terms:** $$ (2a - 1)^6 = 64a^6 - 192a^5 + 240a^4 - 160a^3 + 60a^2 - 12a + 1 $$ 6. **Explanation:** We used the binomial theorem to expand the power of a binomial. Each term involves a binomial coefficient, powers of $2a$, and alternating signs due to $(-1)^k$. This expansion shows the polynomial form of the original expression.