Expand `(2+x+x^(2) )^(3)`.

Answer

To expand the expression \((2 + x + x^2)^3\) using the Binomial Theorem, we can follow these steps:

### Step 1: Identify \(a\), \(b\), and \(n\)
In the expression \((2 + x + x^2)^3\), we can treat \(2\) as \(a\) and \((x + x^2)\) as \(b\). Thus, we have:
- \(a = 2\)
- \(b = x + x^2\)
- \(n = 3\)

### Step 2: Apply the Binomial Theorem
The Binomial Theorem states that:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
For our case, we can write:
\[ (2 + (x + x^2))^3 = \sum_{k=0}^{3} \binom{3}{k} 2^{3-k} (x + x^2)^k \]

### Step 3: Expand the terms
Now we will expand each term in the summation:
1. For \(k = 0\):
\[ \binom{3}{0} 2^{3} (x + x^2)^{0} = 1 \cdot 8 \cdot 1 = 8 \]

2. For \(k = 1\):
\[ \binom{3}{1} 2^{2} (x + x^2)^{1} = 3 \cdot 4 \cdot (x + x^2) = 12(x + x^2) = 12x + 12x^2 \]

3. For \(k = 2\):
\[ \binom{3}{2} 2^{1} (x + x^2)^{2} = 3 \cdot 2 \cdot (x^2 + 2x^3 + x^4) = 6(x^2 + 2x^3 + x^4) = 6x^2 + 12x^3 + 6x^4 \]

4. For \(k = 3\):
\[ \binom{3}{3} 2^{0} (x + x^2)^{3} = 1 \cdot 1 \cdot (x^3 + 3x^4 + 3x^5 + x^6) = x^3 + 3x^4 + 3x^5 + x^6 \]

### Step 4: Combine all the terms
Now we combine all the terms we calculated:
\[ 8 + (12x + 12x^2) + (6x^2 + 12x^3 + 6x^4) + (x^3 + 3x^4 + 3x^5 + x^6) \]

### Step 5: Group like terms
Now we group the like terms:
- Constant term: \(8\)
- Coefficient of \(x\): \(12x\)
- Coefficient of \(x^2\): \(12x^2 + 6x^2 = 18x^2\)
- Coefficient of \(x^3\): \(12x^3 + x^3 = 13x^3\)
- Coefficient of \(x^4\): \(6x^4 + 3x^4 = 9x^4\)
- Coefficient of \(x^5\): \(3x^5\)
- Coefficient of \(x^6\): \(x^6\)

### Final Result
Putting it all together, we have:
\[ (2 + x + x^2)^3 = 8 + 12x + 18x^2 + 13x^3 + 9x^4 + 3x^5 + x^6 \]