If the first three terms in the expansion of `(a + b)^(n)` are 27, 54 and 36 respectively, then find a, b and n.

Answer

To solve the problem, we need to find the values of \(a\), \(b\), and \(n\) given that the first three terms in the expansion of \((a + b)^n\) are 27, 54, and 36 respectively.

### Step 1: Write the first three terms of the binomial expansion
The first three terms of the expansion of \((a + b)^n\) are:
1. \(T_0 = \binom{n}{0} a^n b^0 = a^n\)
2. \(T_1 = \binom{n}{1} a^{n-1} b^1 = n a^{n-1} b\)
3. \(T_2 = \binom{n}{2} a^{n-2} b^2 = \frac{n(n-1)}{2} a^{n-2} b^2\)

Given:
- \(T_0 = a^n = 27\)
- \(T_1 = n a^{n-1} b = 54\)
- \(T_2 = \frac{n(n-1)}{2} a^{n-2} b^2 = 36\)

### Step 2: Set up equations based on the given terms
From the first term:
\[ a^n = 27 \quad \text{(1)} \]

From the second term:
\[ n a^{n-1} b = 54 \quad \text{(2)} \]

From the third term:
\[ \frac{n(n-1)}{2} a^{n-2} b^2 = 36 \quad \text{(3)} \]

### Step 3: Express \(b\) in terms of \(a\) and \(n\)
From equation (1), we can express \(a\) as:
\[ a = 27^{1/n} \quad \text{(4)} \]

Substituting equation (4) into equation (2):
\[ n (27^{1/n})^{n-1} b = 54 \]
This simplifies to:
\[ n \cdot 27^{(n-1)/n} b = 54 \]
Thus,
\[ b = \frac{54}{n \cdot 27^{(n-1)/n}} \quad \text{(5)} \]

### Step 4: Substitute \(b\) into equation (3)
Now, substituting \(b\) from equation (5) into equation (3):
\[ \frac{n(n-1)}{2} (27^{1/n})^{n-2} \left(\frac{54}{n \cdot 27^{(n-1)/n}}\right)^2 = 36 \]
This simplifies to:
\[ \frac{n(n-1)}{2} \cdot 27^{(n-2)/n} \cdot \frac{2916}{n^2 \cdot 27^{(n-1)/n}} = 36 \]
Thus,
\[ \frac{(n-1) \cdot 2916}{2n} \cdot 27^{-1/n} = 36 \]

### Step 5: Solve for \(n\)
Cross-multiplying gives:
\[ (n-1) \cdot 2916 = 72n \cdot 27^{1/n} \]
Now, we can try different integer values for \(n\) to find a suitable solution.

### Step 6: Testing \(n = 3\)
Let’s test \(n = 3\):
\[ a = 27^{1/3} = 3 \]
Substituting \(n = 3\) into equation (2):
\[ 3 \cdot 3^{2} b = 54 \implies 27b = 54 \implies b = 2 \]

### Step 7: Verify with equation (3)
Now, substituting \(n = 3\), \(a = 3\), and \(b = 2\) into equation (3):
\[ \frac{3(3-1)}{2} (3^{1}) (2^2) = 36 \]
This simplifies to:
\[ \frac{3 \cdot 2}{2} \cdot 3 \cdot 4 = 36 \implies 3 \cdot 3 \cdot 4 = 36 \]
This is correct.

### Final Values
Thus, the values are:
- \(a = 3\)
- \(b = 2\)
- \(n = 3\)