If `f(x)=x^(3)+3x^(2)+3x`, then `f(x+1)` is equal to

Options

A. `x^(3)+6x^(2)+12x+7`
B. `x^(3)+4x^(2)+5`
C. `3x^(3)+12x^(2)+4x+1`
D. `2x^(3)+6x^(2)+3`

Answer

To find \( f(x+1) \) for the function \( f(x) = x^3 + 3x^2 + 3x \), we will follow these steps:

### Step 1: Substitute \( x+1 \) into the function
We start by substituting \( x+1 \) into the function \( f(x) \):
\[ f(x+1) = (x+1)^3 + 3(x+1)^2 + 3(x+1) \]

### Step 2: Expand \( (x+1)^3 \)
Using the binomial expansion, we have:
\[ (x+1)^3 = x^3 + 3x^2 + 3x + 1 \]

### Step 3: Expand \( 3(x+1)^2 \)
First, we expand \( (x+1)^2 \):
\[ (x+1)^2 = x^2 + 2x + 1 \]
Now, multiply by 3:
\[ 3(x+1)^2 = 3(x^2 + 2x + 1) = 3x^2 + 6x + 3 \]

### Step 4: Expand \( 3(x+1) \)
Now, we expand \( 3(x+1) \):
\[ 3(x+1) = 3x + 3 \]

### Step 5: Combine all the expanded parts
Now we combine all the expanded parts:
\[ f(x+1) = (x^3 + 3x^2 + 3x + 1) + (3x^2 + 6x + 3) + (3x + 3) \]

### Step 6: Simplify the expression
Now, we combine like terms:
- \( x^3 \) terms: \( x^3 \)
- \( x^2 \) terms: \( 3x^2 + 3x^2 = 6x^2 \)
- \( x \) terms: \( 3x + 6x + 3x = 12x \)
- Constant terms: \( 1 + 3 + 3 = 7 \)

Putting it all together, we get:
\[ f(x+1) = x^3 + 6x^2 + 12x + 7 \]

### Final Answer
Thus, the value of \( f(x+1) \) is:
\[ \boxed{x^3 + 6x^2 + 12x + 7} \]