Let A;G;H be the arithmetic; geometric and harmonic means between three given no. a;b;c then the equation having a;b;c as its root is

Explanation

To find the equation having \( a, b, c \) as its roots, where \( A, G, H \) are the arithmetic, geometric, and harmonic means of the numbers \( a, b, c \), we can follow these steps:

### Step 1: Define the Means
1. **Arithmetic Mean (A)**:
\[ A = \frac{a + b + c}{3} \]
Therefore, we have:
\[ a + b + c = 3A \]

2. **Geometric Mean (G)**:
\[ G = \sqrt[3]{abc} \]
Hence, we can express \( abc \) as:
\[ abc = G^3 \]

3. **Harmonic Mean (H)**:
\[ H = \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} = \frac{3abc}{ab + ac + bc} \]
Rearranging gives:
\[ ab + ac + bc = \frac{3abc}{H} \]

### Step 2: Form the Cubic Equation
Using Vieta's formulas, we know that for a cubic equation \( x^3 - (sum \ of \ roots)x^2 + (sum \ of \ products \ of \ roots \ taken \ two \ at \ a \ time)x - (product \ of \ roots) = 0 \), we can substitute the values we found:

- **Sum of the roots**: \( a + b + c = 3A \)
- **Sum of the products of the roots taken two at a time**: \( ab + ac + bc = \frac{3abc}{H} \)
- **Product of the roots**: \( abc = G^3 \)

Thus, the cubic equation becomes:
\[ x^3 - (3A)x^2 + \left(\frac{3abc}{H}\right)x - abc = 0 \]

Substituting \( abc = G^3 \) into the equation gives:
\[ x^3 - (3A)x^2 + \left(\frac{3G^3}{H}\right)x - G^3 = 0 \]

### Final Equation
The final cubic equation having \( a, b, c \) as its roots is:
\[ x^3 - 3Ax^2 + \frac{3G^3}{H}x - G^3 = 0 \]