Find the condition that `x^3-px^2+qx-r=0` may have the roots in G.P .

Answer

To find the condition that the cubic equation \( x^3 - px^2 + qx - r = 0 \) may have roots in Geometric Progression (G.P.), we can follow these steps:

### Step 1: Define the Roots
Let the roots of the equation be \( \alpha, \beta, \gamma \). Since the roots are in G.P., we can express them as:
\[ \alpha = \frac{a}{r}, \quad \beta = a, \quad \gamma = ar \]
where \( a \) is the middle term and \( r \) is the common ratio.

### Step 2: Use Vieta's Formulas
According to Vieta's formulas for a cubic equation \( x^3 + bx^2 + cx + d = 0 \):
1. The sum of the roots \( \alpha + \beta + \gamma = -\frac{b}{a} \)
2. The sum of the products of the roots taken two at a time \( \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} \)
3. The product of the roots \( \alpha\beta\gamma = -\frac{d}{a} \)

For our equation \( x^3 - px^2 + qx - r = 0 \), we have:
- \( a = 1 \)
- \( b = -p \)
- \( c = q \)
- \( d = -r \)

### Step 3: Calculate the Sum of the Roots
Using Vieta's formulas:
\[ \alpha + \beta + \gamma = \frac{a}{r} + a + ar = p \]
Factoring out \( a \):
\[ a \left( \frac{1}{r} + 1 + r \right) = p \]

### Step 4: Calculate the Product of the Roots
The product of the roots:
\[ \alpha \beta \gamma = \left(\frac{a}{r}\right) \cdot a \cdot (ar) = a^3 = r \]

### Step 5: Calculate the Sum of the Products of the Roots
The sum of the products of the roots taken two at a time:
\[ \alpha \beta + \beta \gamma + \gamma \alpha = \left(\frac{a}{r}\right)a + a(ar) + \left(\frac{a}{r}\right)(ar) = \frac{a^2}{r} + a^2 + \frac{a^2}{r} = a^2 \left( \frac{1}{r} + 1 + r \right) = q \]

### Step 6: Relate the Equations
From the product of the roots, we have:
\[ a^3 = r \]
From the sum of the roots, we have:
\[ a \left( \frac{1}{r} + 1 + r \right) = p \]
From the sum of the products of the roots:
\[ a^2 \left( \frac{1}{r} + 1 + r \right) = q \]

### Step 7: Substitute and Simplify
Substituting \( a^3 = r \) into the equations gives us:
\[ \frac{1}{r} + 1 + r = \frac{p}{a} \]
and
\[ \frac{1}{r} + 1 + r = \frac{q}{a^2} \]

### Step 8: Set the Conditions
From the equations:
1. \( a = \frac{q}{p} \)
2. \( r = \left(\frac{q}{p}\right)^3 \)

Thus, we derive the condition:
\[ q^3 = rp^2 \]

### Final Condition
The condition that the cubic equation \( x^3 - px^2 + qx - r = 0 \) may have roots in G.P. is:
\[ q^3 = rp^2 \]