Find the condition that `x^3-px^2+qx-r=0` may have the sum of its roots zero .

Answer

To find the condition that the cubic equation \( x^3 - px^2 + qx - r = 0 \) has the sum of its roots equal to zero, we can follow these steps:

### Step 1: Understand the sum of the roots
For a cubic equation of the form \( ax^3 + bx^2 + cx + d = 0 \), the sum of the roots (let's denote them as \( \alpha, \beta, \gamma \)) can be given by the formula:
\[ \alpha + \beta + \gamma = -\frac{b}{a} \]
In our case, the equation is \( x^3 - px^2 + qx - r = 0 \), where \( a = 1 \), \( b = -p \), \( c = q \), and \( d = -r \). Therefore, the sum of the roots becomes:
\[ \alpha + \beta + \gamma = -\frac{-p}{1} = p \]

### Step 2: Set the condition for the sum of the roots
We want the sum of the roots to be zero:
\[ \alpha + \beta + \gamma = 0 \]
This implies:
\[ p = 0 \]

### Step 3: Express one root in terms of the others
Assuming \( \alpha + \beta = 0 \), we can denote \( \beta = -\alpha \). Thus, we have:
\[ \gamma = p \]

### Step 4: Substitute \( \gamma \) into the original equation
We now substitute \( \gamma \) into the original cubic equation:
\[ \gamma^3 - p\gamma^2 + q\gamma - r = 0 \]
Substituting \( \gamma = p \):
\[ p^3 - p(p^2) + qp - r = 0 \]

### Step 5: Simplify the equation
This simplifies to:
\[ p^3 - p^3 + qp - r = 0 \]
Thus:
\[ qp - r = 0 \]
Rearranging gives us the condition:
\[ qp = r \]

### Final Condition
The condition that the cubic equation \( x^3 - px^2 + qx - r = 0 \) has the sum of its roots equal to zero is:
\[ r = qp \]

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