Write down and simplify
5th term in `(3x - 4y)^7`

Answer

To find the fifth term in the expansion of \((3x - 4y)^7\), we will use the Binomial Theorem. The Binomial Theorem states that:

\[ (a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r \]

Where:
- \(\binom{n}{r}\) is the binomial coefficient, calculated as \(\frac{n!}{r!(n-r)!}\)
- \(a\) and \(b\) are the terms in the binomial
- \(n\) is the exponent
- \(r\) is the term index (starting from 0)

### Step 1: Identify the parameters
In our case:
- \(a = 3x\)
- \(b = -4y\)
- \(n = 7\)

### Step 2: Determine the term index
To find the fifth term, we need to set \(r = 4\) (since the first term corresponds to \(r = 0\)).

### Step 3: Use the Binomial Theorem formula
The \( (r+1) \)th term (which is the 5th term when \(r=4\)) is given by:

\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]

Substituting the values we have:

\[ T_5 = \binom{7}{4} (3x)^{7-4} (-4y)^4 \]

### Step 4: Calculate the binomial coefficient
Calculate \(\binom{7}{4}\):

\[ \binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7!}{4! \cdot 3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]

### Step 5: Calculate \(a^{n-r}\) and \(b^r\)
Now calculate \( (3x)^{3} \) and \( (-4y)^{4} \):

\[ (3x)^3 = 27x^3 \]
\[ (-4y)^4 = 256y^4 \]

### Step 6: Combine all parts
Now substitute back into the term:

\[ T_5 = 35 \cdot 27x^3 \cdot 256y^4 \]

### Step 7: Simplify the expression
Now, multiply the constants together:

\[ T_5 = 35 \cdot 27 \cdot 256 \cdot x^3 \cdot y^4 \]

Calculating \(35 \cdot 27 \cdot 256\):

\[ 35 \cdot 27 = 945 \]
\[ 945 \cdot 256 = 241920 \]

Thus, we have:

\[ T_5 = 241920 x^3 y^4 \]

### Final Answer
The fifth term in the expansion of \((3x - 4y)^7\) is:

\[ \boxed{241920 x^3 y^4} \]