If one root of the equation `ax^(2)+bx+c=0` be the square of the other, then

Options

A. `a^(3)+bc (b+c)=3abc`
B. `b^(3)+ac (a+c)=3abc`
C. `c^(3)+ab (a+b)=3abc`
D. none

Answer

To solve the problem where one root of the quadratic equation \( ax^2 + bx + c = 0 \) is the square of the other root, we can follow these steps:

### Step 1: Define the Roots
Let the roots of the quadratic equation be \( \alpha \) and \( \alpha^2 \).

### Step 2: Use the Sum of Roots
According to Vieta's formulas, the sum of the roots is given by:
\[ \alpha + \alpha^2 = -\frac{b}{a} \]

### Step 3: Use the Product of Roots
The product of the roots is given by:
\[ \alpha \cdot \alpha^2 = \alpha^3 = \frac{c}{a} \]

### Step 4: Rearranging the Sum of Roots
From the sum of roots equation, we can express \( \alpha^2 + \alpha \):
\[ \alpha^2 + \alpha = -\frac{b}{a} \]

### Step 5: Cube the Sum of Roots
Now, we cube both sides of the equation:
\[ (\alpha^2 + \alpha)^3 = \left(-\frac{b}{a}\right)^3 \]

### Step 6: Expand the Left Side
Using the binomial expansion:
\[ \alpha^6 + 3\alpha^4 \cdot \alpha + 3\alpha^2 \cdot \alpha^2 + \alpha^3 = -\frac{b^3}{a^3} \]
This simplifies to:
\[ \alpha^6 + 3\alpha^3 + 3\alpha^4 = -\frac{b^3}{a^3} \]

### Step 7: Substitute for \(\alpha^3\)
From the product of roots, we know \( \alpha^3 = \frac{c}{a} \). Substitute this into the equation:
\[ \alpha^6 + 3\left(\frac{c}{a}\right) + 3\alpha^4 = -\frac{b^3}{a^3} \]

### Step 8: Express \(\alpha^4\) in terms of \(\alpha\)
We can express \( \alpha^4 \) as \( \alpha^2 \cdot \alpha^2 = \alpha^2 \cdot \left(-\frac{b}{a} - \alpha\right) \). However, for simplicity, we can use the earlier derived equations to eliminate \(\alpha\).

### Step 9: Multiply through by \( a^3 \)
To eliminate the denominators, multiply through by \( a^3 \):
\[ a^3 \alpha^6 + 3a^2c + 3a^3\alpha^4 = -b^3 \]

### Step 10: Rearranging the Equation
Rearranging gives us:
\[ a^3 \alpha^6 + 3a^2c + 3a^3\alpha^4 + b^3 = 0 \]

### Final Step: Collect Terms
This leads us to the final relation:
\[ b^3 + 3abc + ac^2 = 0 \]

### Conclusion
Thus, if one root of the equation \( ax^2 + bx + c = 0 \) is the square of the other, we arrive at the relation:
\[ b^3 + 3abc + ac^2 = 0 \]