Find the 3rd term the end in the expansion of `(2-3x)^(8)`

Explanation

To find the third term from the end in the expansion of \((2 - 3x)^8\), we can follow these steps:

### Step 1: Identify the general term in the binomial expansion
The general term \(T_r\) in the binomial expansion of \((a + b)^n\) is given by:
\[ T_r = \binom{n}{r-1} a^{n - (r-1)} b^{r-1} \]
where \(n\) is the exponent, \(a\) is the first term, \(b\) is the second term, and \(r\) is the term number.

### Step 2: Determine the parameters for our specific case
In our case, we have:
- \(a = 2\)
- \(b = -3x\)
- \(n = 8\)

### Step 3: Find the term number from the end
To find the third term from the end, we need to find the corresponding term from the beginning. The \(r\)-th term from the end is given by:
\[ T_{n - r + 1} \]
For the third term from the end:
\[ r = 3 \implies T_{8 - 3 + 1} = T_6 \]

### Step 4: Calculate the 6th term in the expansion
Using the general term formula, we find \(T_6\):
\[ T_6 = \binom{8}{6-1} (2)^{8 - (6-1)} (-3x)^{6-1} \]
This simplifies to:
\[ T_6 = \binom{8}{5} (2)^{8 - 5} (-3x)^{5} \]

### Step 5: Calculate the binomial coefficient and powers
Now we calculate each part:
1. \(\binom{8}{5} = \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56\)
2. \(2^{8-5} = 2^3 = 8\)
3. \((-3x)^5 = -243x^5\)

### Step 6: Combine the results
Now, substituting these values back into the term:
\[ T_6 = 56 \cdot 8 \cdot (-243x^5) \]
Calculating this gives:
\[ T_6 = 56 \cdot 8 \cdot -243 = -108864x^5 \]

### Final Result
Thus, the third term from the end in the expansion of \((2 - 3x)^8\) is:
\[ -108864x^5 \]