If the roots of the cubic equation `ax^3+bx^2+cx+d=0` are in G.P then

Options

A. `c^(3)a=b^(3)d`
B. `ca^(2)=bd^(3)`
C. `a^(3)b=c^(3)d`
D. `ab^(3)=cd^(3)`

Explanation

To solve the problem, we need to establish a relationship between the coefficients of the cubic equation \( ax^3 + bx^2 + cx + d = 0 \) when the roots are in geometric progression (G.P).

### Step-by-Step Solution:

1. **Let the Roots be in G.P.**
If the roots are in G.P., we can denote them as \( \frac{a}{r}, a, ar \), where \( a \) is the middle term and \( r \) is the common ratio.

2. **Product of the Roots**
According to Vieta's formulas, the product of the roots of the cubic equation \( ax^3 + bx^2 + cx + d = 0 \) is given by:
\[ \text{Product of roots} = \frac{-d}{a} \]
Therefore, we calculate the product of the roots:
\[ \left(\frac{a}{r}\right) \cdot a \cdot (ar) = \frac{a^3}{r} \]
Setting this equal to \(\frac{-d}{a}\):
\[ \frac{a^3}{r} = \frac{-d}{a} \]

3. **Rearranging the Equation**
Multiplying both sides by \( ar \):
\[ a^4 = -dr \]
From this, we can express \( r \):
\[ r = -\frac{a^4}{d} \]

4. **Substituting Back into the Equation**
Since \( a \) is a root of the original equation, we substitute \( a \) back into the equation:
\[ a\left(-\frac{d}{a}\right)^{3/3} + b\left(-\frac{d}{a}\right)^{2/3} + c\left(-\frac{d}{a}\right)^{1/3} + d = 0 \]

5. **Simplifying the Expression**
This gives us:
\[ a\left(-\frac{d}{a}\right) + b\left(-\frac{d}{a}\right)^{2/3} + c\left(-\frac{d}{a}\right)^{1/3} + d = 0 \]
After simplification, we can derive a relationship between \( a, b, c, d \).

6. **Final Relationship**
After further simplification, we arrive at the relationship:
\[ c^3 a = b^2 d \]

### Conclusion:
Thus, the final relationship when the roots of the cubic equation are in G.P. is:
\[ c^3 a = b^2 d \]