Using binomial theorem, expand each of the following: `(2x-3)^(6)`

Answer

To expand \((2x - 3)^6\) using the Binomial Theorem, we follow these steps:

### Step 1: Identify the Binomial Theorem Formula
The Binomial Theorem states that:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
where \(\binom{n}{k}\) is the binomial coefficient.

### Step 2: Assign Values
In our case, we have:
- \(a = 2x\)
- \(b = -3\)
- \(n = 6\)

### Step 3: Write the Expansion
Using the Binomial Theorem, we can write the expansion as:
\[ (2x - 3)^6 = \sum_{k=0}^{6} \binom{6}{k} (2x)^{6-k} (-3)^k \]

### Step 4: Calculate Each Term
Now we will calculate each term from \(k = 0\) to \(k = 6\):

1. **For \(k = 0\)**:
\[ \binom{6}{0} (2x)^6 (-3)^0 = 1 \cdot 64x^6 \cdot 1 = 64x^6 \]

2. **For \(k = 1\)**:
\[ \binom{6}{1} (2x)^5 (-3)^1 = 6 \cdot 32x^5 \cdot (-3) = -576x^5 \]

3. **For \(k = 2\)**:
\[ \binom{6}{2} (2x)^4 (-3)^2 = 15 \cdot 16x^4 \cdot 9 = 2160x^4 \]

4. **For \(k = 3\)**:
\[ \binom{6}{3} (2x)^3 (-3)^3 = 20 \cdot 8x^3 \cdot (-27) = -4320x^3 \]

5. **For \(k = 4\)**:
\[ \binom{6}{4} (2x)^2 (-3)^4 = 15 \cdot 4x^2 \cdot 81 = 4860x^2 \]

6. **For \(k = 5\)**:
\[ \binom{6}{5} (2x)^1 (-3)^5 = 6 \cdot 2x \cdot (-243) = -2916x \]

7. **For \(k = 6\)**:
\[ \binom{6}{6} (2x)^0 (-3)^6 = 1 \cdot 1 \cdot 729 = 729 \]

### Step 5: Combine All Terms
Now, we combine all the terms we calculated:
\[ (2x - 3)^6 = 64x^6 - 576x^5 + 2160x^4 - 4320x^3 + 4860x^2 - 2916x + 729 \]

### Final Answer
Thus, the expansion of \((2x - 3)^6\) is:
\[ \boxed{64x^6 - 576x^5 + 2160x^4 - 4320x^3 + 4860x^2 - 2916x + 729} \]