To expand and simplify the binomial $(3a - 2b)^5$, we use the Binomial Theorem, which states:

$(x + y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k} y^k$

Here:

• $n = 5$ (the power),
• $x = 3a$ and $y = -2b$.

The general term is given as:

$T_k = \binom{5}{k} (3a)^{5-k} (-2b)^k$

Now we calculate each term by substituting values of $k$ from 0 to 5:

Step-by-step Expansion:

1. For $k = 0$: $T_0 = \binom{5}{0} (3a)^5 (-2b)^0 = 1 \cdot 243a^5 \cdot 1 = 243a^5$

2. For $k = 1$: $T_1 = \binom{5}{1} (3a)^4 (-2b)^1 = 5 \cdot 81a^4 \cdot (-2b) = -810a^4b$

3. For $k = 2$: $T_2 = \binom{5}{2} (3a)^3 (-2b)^2 = 10 \cdot 27a^3 \cdot 4b^2 = 1080a^3b^2$

4. For $k = 3$: $T_3 = \binom{5}{3} (3a)^2 (-2b)^3 = 10 \cdot 9a^2 \cdot (-8b^3) = -720a^2b^3$

5. For $k = 4$: $T_4 = \binom{5}{4} (3a)^1 (-2b)^4 = 5 \cdot 3a \cdot 16b^4 = 240ab^4$

6. For $k = 5$: $T_5 = \binom{5}{5} (3a)^0 (-2b)^5 = 1 \cdot 1 \cdot (-32b^5) = -32b^5$

Combine All Terms:

Adding all terms together, we get:

$(3a - 2b)^5 = 243a^5 - 810a^4b + 1080a^3b^2 - 720a^2b^3 + 240ab^4 - 32b^5$

Final Answer:

$\boxed{243a^5 - 810a^4b + 1080a^3b^2 - 720a^2b^3 + 240ab^4 - 32b^5}$

Let me know if you need further clarification or steps explained!

Related Questions:

1. How does the binomial theorem work for negative terms?
2. Can you expand $(x + y)^6$ step-by-step?
3. How do you find the general term of a binomial expansion?
4. What is Pascal’s Triangle and how is it related to binomial coefficients?
5. How can the binomial theorem be applied to approximations?

Tip:

The binomial theorem is a quick way to expand powers of binomials without direct multiplication.