If the roots of the equation `x^(3) + bx^(2) + cx - 1 = 0` form an increasing G.P., then b belongs to which interval ?

Options

A. `b+c=0`
B. `b epsilon(-oo,-3)`
C. one of the roots is 1
D. one root is smaller than one and one root is more than one

Explanation

To determine the interval in which \( b \) belongs, given that the roots of the equation \( x^3 + bx^2 + cx - 1 = 0 \) form an increasing geometric progression (G.P.), we can follow these steps:

### Step 1: Define the Roots
Let the roots of the polynomial be \( \frac{a}{r}, a, ar \), where \( r > 1 \) (since the roots are in increasing order).

### Step 2: Sum of the Roots
Using Vieta's formulas, the sum of the roots can be expressed as:
\[ \frac{a}{r} + a + ar = -b \]
This can be simplified to:
\[ \frac{a}{r} + a + ar = -b \quad \text{(Equation 1)} \]

### Step 3: Product of the Roots
The product of the roots is given by:
\[ \left(\frac{a}{r}\right) \cdot a \cdot (ar) = 1 \]
This simplifies to:
\[ \frac{a^3}{r} = 1 \implies a^3 = r \]

### Step 4: Substitute \( a \) in Equation 1
Substituting \( a = 1 \) (since \( a^3 = r \) implies \( a = 1 \) when \( r = 1 \)):
\[ \frac{1}{r} + 1 + r = -b \]
This simplifies to:
\[ \frac{1 + r + r^2}{r} = -b \]
Multiplying through by \( r \) gives:
\[ 1 + r + r^2 = -br \]
Rearranging leads to:
\[ r^2 + r + b r + 1 = 0 \]
This can be rewritten as:
\[ r^2 + (b + 1)r + 1 = 0 \quad \text{(Equation 2)} \]

### Step 5: Discriminant Condition
For \( r \) to be real, the discriminant of Equation 2 must be greater than zero:
\[ (b + 1)^2 - 4 \cdot 1 \cdot 1 > 0 \]
This simplifies to:
\[ (b + 1)^2 - 4 > 0 \]
\[ (b + 1)^2 > 4 \]

### Step 6: Solve the Inequality
Taking square roots gives us two cases:
1. \( b + 1 > 2 \) or
2. \( b + 1 < -2 \)

From the first case:
\[ b > 1 \]

From the second case:
\[ b < -3 \]

### Step 7: Conclusion
Thus, \( b \) belongs to the intervals:
\[ b < -3 \quad \text{or} \quad b > 1 \]

### Final Answer
The intervals for \( b \) are:
\[ (-\infty, -3) \cup (1, \infty) \]