Find the middle terms are are there in the expansion of `(3a-5b)^6 ?`

Answer

To find the middle term in the expansion of \((3a - 5b)^6\), we can follow these steps:

### Step 1: Identify the value of \(n\)
In the expression \((3a - 5b)^6\), we see that \(n = 6\).

**Hint:** The exponent in the binomial expression indicates the value of \(n\).

### Step 2: Determine the middle term
Since \(n\) is even, the middle term can be found using the formula for the middle term in a binomial expansion, which is given by:
\[ \text{Middle Term} = \left(\frac{n}{2} + 1\right)^{th} \text{ term} \]
Here, \(n = 6\), so:
\[ \frac{6}{2} + 1 = 3 + 1 = 4 \]
Thus, the middle term is the 4th term.

**Hint:** For an even \(n\), the middle term is the \(\left(\frac{n}{2} + 1\right)^{th}\) term.

### Step 3: Use the formula for the \(r^{th}\) term
The \(r^{th}\) term in the expansion of \((x + y)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} x^{n-r} y^r \]
For our case, we need the 4th term, which corresponds to \(r = 3\) (since \(T_{r+1} = T_4\)).

**Hint:** Remember that \(r\) starts from 0, so the 4th term corresponds to \(r = 3\).

### Step 4: Substitute values into the formula
Substituting \(n = 6\), \(r = 3\), \(x = 3a\), and \(y = -5b\):
\[ T_4 = \binom{6}{3} (3a)^{6-3} (-5b)^3 \]

**Hint:** Make sure to substitute correctly for \(x\) and \(y\).

### Step 5: Calculate the binomial coefficient
Calculate \(\binom{6}{3}\):
\[ \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]

**Hint:** The binomial coefficient can be calculated using factorials.

### Step 6: Calculate the powers
Now calculate the powers:
\[ (3a)^{3} = 27a^3 \]
\[ (-5b)^{3} = -125b^3 \]

**Hint:** Remember that raising a negative number to an odd power results in a negative outcome.

### Step 7: Combine all parts
Now substitute back into the term:
\[ T_4 = 20 \cdot 27a^3 \cdot (-125b^3) \]

### Step 8: Simplify the expression
Calculate:
\[ T_4 = 20 \cdot 27 \cdot (-125) \cdot a^3 \cdot b^3 \]
Calculating \(20 \cdot 27 = 540\) and \(540 \cdot (-125) = -67500\):
\[ T_4 = -67500 a^3 b^3 \]

### Final Answer
The middle term in the expansion of \((3a - 5b)^6\) is:
\[ -67500 a^3 b^3 \]