If `x^3-3x^2+3x+7=(x+1)(a x^2+b x+c),\ ` then `a+b+c=`

Options

A. 4
B. 12
C. `-10`
D. 3

Answer

To solve the equation \( x^3 - 3x^2 + 3x + 7 = (x + 1)(ax^2 + bx + c) \), we will follow these steps:

### Step 1: Expand the Right Side
We start by expanding the right side of the equation:

\[ (x + 1)(ax^2 + bx + c) = ax^3 + bx^2 + cx + ax^2 + bx + c \]

Combining like terms, we get:

\[ = ax^3 + (b + a)x^2 + (c + b)x + c \]

### Step 2: Set Coefficients Equal
Now, we will compare the coefficients of the left side \( x^3 - 3x^2 + 3x + 7 \) with the expanded form:

1. Coefficient of \( x^3 \): \( a = 1 \)
2. Coefficient of \( x^2 \): \( b + a = -3 \)
3. Coefficient of \( x \): \( c + b = 3 \)
4. Constant term: \( c = 7 \)

### Step 3: Solve for \( a, b, c \)
From the first equation, we have:

\[ a = 1 \]

Substituting \( a = 1 \) into the second equation:

\[ b + 1 = -3 \implies b = -4 \]

Now substituting \( b = -4 \) into the third equation:

\[ c - 4 = 3 \implies c = 7 \]

### Step 4: Calculate \( a + b + c \)
Now we can find \( a + b + c \):

\[ a + b + c = 1 - 4 + 7 = 4 \]

### Final Answer
Thus, the value of \( a + b + c \) is:

\[ \boxed{4} \]