If the roots of the cubic `x ^(3) +ax ^(2) + bx +c=0` are three consecutive positive integers, then the value of `(a ^(2))/(b +1) =`

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to find the value of \(\frac{a^2}{b + 1}\) given that the roots of the cubic equation \(x^3 + ax^2 + bx + c = 0\) are three consecutive positive integers.

### Step-by-Step Solution:

1. **Define the Roots:**
Let's denote the three consecutive positive integers as \( \alpha - 1, \alpha, \alpha + 1\).

2. **Sum of the Roots:**
According to Vieta's formulas, the sum of the roots is given by:
\[ (\alpha - 1) + \alpha + (\alpha + 1) = 3\alpha \]
This sum is equal to \(-a\) (the coefficient of \(x^2\) with a negative sign). Therefore, we have:
\[ 3\alpha = -a \implies a = -3\alpha \]

3. **Calculate \(a^2\):**
Now, we can find \(a^2\):
\[ a^2 = (-3\alpha)^2 = 9\alpha^2 \]

4. **Product of the Roots:**
The product of the roots taken two at a time is given by:
\[ (\alpha - 1)\alpha + \alpha(\alpha + 1) + (\alpha - 1)(\alpha + 1) \]
Simplifying each term:
- \((\alpha - 1)\alpha = \alpha^2 - \alpha\)
- \(\alpha(\alpha + 1) = \alpha^2 + \alpha\)
- \((\alpha - 1)(\alpha + 1) = \alpha^2 - 1\)

Adding these together:
\[ (\alpha^2 - \alpha) + (\alpha^2 + \alpha) + (\alpha^2 - 1) = 3\alpha^2 - 1 \]
According to Vieta's formulas, this sum is equal to \(b\). Therefore:
\[ b = 3\alpha^2 - 1 \]

5. **Calculate \(b + 1\):**
Now, we can find \(b + 1\):
\[ b + 1 = (3\alpha^2 - 1) + 1 = 3\alpha^2 \]

6. **Calculate \(\frac{a^2}{b + 1}\):**
Now we can substitute \(a^2\) and \(b + 1\) into the expression:
\[ \frac{a^2}{b + 1} = \frac{9\alpha^2}{3\alpha^2} = 3 \]

### Final Result:
Thus, the value of \(\frac{a^2}{b + 1}\) is \(3\).