Let a,b,c be three roots of the equation `x^3+x^2-333x-1002=0` then `((suma^3)-2suma)` is

Options

A. 2008
B. 2000
C. 2006
D. 2002

Answer

To solve the problem, we need to find the value of \( S = (a^3 + b^3 + c^3) - 2(a + b + c) \) where \( a, b, c \) are the roots of the polynomial \( x^3 + x^2 - 333x - 1002 = 0 \).

### Step 1: Identify the coefficients of the polynomial
The polynomial is given as:
\[ x^3 + x^2 - 333x - 1002 = 0 \]
From this, we can identify:
- \( a = 1 \) (coefficient of \( x^3 \))
- \( b = 1 \) (coefficient of \( x^2 \))
- \( c = -333 \) (coefficient of \( x \))
- \( d = -1002 \) (constant term)

### Step 2: Use Vieta's formulas to find the sums of the roots
According to Vieta's formulas:
1. The sum of the roots \( a + b + c = -\frac{b}{a} = -\frac{1}{1} = -1 \)
2. The sum of the products of the roots taken two at a time \( ab + ac + bc = \frac{c}{a} = -333 \)
3. The product of the roots \( abc = -\frac{d}{a} = -\frac{-1002}{1} = 1002 \)

### Step 3: Calculate \( a^3 + b^3 + c^3 \)
We can use the identity:
\[ a^3 + b^3 + c^3 = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) + 3abc \]
First, we need to find \( a^2 + b^2 + c^2 \):
\[ a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + ac + bc) \]
Substituting the known values:
\[ a^2 + b^2 + c^2 = (-1)^2 - 2(-333) = 1 + 666 = 667 \]
Now, substituting back into the identity:
\[ a^3 + b^3 + c^3 = (-1)(667 - (-333)) + 3(1002) \]
Calculating the expression inside the parentheses:
\[ 667 + 333 = 1000 \]
Thus:
\[ a^3 + b^3 + c^3 = -1 \times 1000 + 3006 = -1000 + 3006 = 2006 \]

### Step 4: Calculate \( S = (a^3 + b^3 + c^3) - 2(a + b + c) \)
Now we can substitute the values we have found:
\[ S = 2006 - 2(-1) = 2006 + 2 = 2008 \]

### Final Answer
Thus, the value of \( S \) is:
\[ \boxed{2008} \]