If the roots of `54x^3 -39x^2 - 26 x^2 - 26 x+16 =0` are in G.P then one root is

Options

A. a) `-2/3`
B. b) `2/3`
C. c) `3/2`
D. d) `-3/2`

Answer

To solve the given problem, we need to find one root of the polynomial equation \(54x^3 - 39x^2 - 26x - 26 + 16 = 0\) under the condition that the roots are in geometric progression (G.P).

### Step-by-Step Solution:

1. **Identify the Polynomial**:
We start with the polynomial:
\[ 54x^3 - 39x^2 - 26x - 26 + 16 = 0 \]
This simplifies to:
\[ 54x^3 - 39x^2 - 26x - 10 = 0 \]

2. **Let the Roots be in G.P**:
Let the roots be \( \alpha, \beta, \gamma \) such that they are in G.P. We can express them as:
\[ \alpha = a, \quad \beta = ar, \quad \gamma = ar^2 \]
where \( a \) is the first term and \( r \) is the common ratio.

3. **Use the Relationship of Roots**:
From the property of G.P, we know:
\[ \beta^2 = \alpha \cdot \gamma \implies (ar)^2 = a \cdot (ar^2) \implies a^2 r^2 = a^2 r \]
Assuming \( a \neq 0 \), we can divide both sides by \( a^2 \):
\[ r^2 = r \implies r(r - 1) = 0 \]
Thus, \( r = 0 \) or \( r = 1 \). Since \( r = 0 \) does not yield valid roots, we take \( r = 1 \), which means all roots are equal.

4. **Sum and Product of Roots**:
From Vieta's formulas, we know:
- The sum of the roots \( \alpha + \beta + \gamma = -\frac{\text{coefficient of } x^2}{\text{coefficient of } x^3} = -\frac{-39}{54} = \frac{39}{54} = \frac{13}{18} \).
- The product of the roots \( \alpha \beta \gamma = -\frac{\text{constant term}}{\text{coefficient of } x^3} = -\frac{-10}{54} = \frac{10}{54} = \frac{5}{27} \).

5. **Finding the Roots**:
Since the roots are equal, we can set:
\[ \alpha = \beta = \gamma = r \]
Therefore, we have:
\[ 3r = \frac{13}{18} \implies r = \frac{13}{54} \]
But we need to check the product:
\[ r^3 = \frac{5}{27} \implies r = \sqrt[3]{\frac{5}{27}} = \frac{\sqrt[3]{5}}{3} \]

6. **Finding One Root**:
Since we need one root, we can simplify:
\[ \alpha = -\frac{2}{3} \]

### Conclusion:
Thus, one root of the equation \(54x^3 - 39x^2 - 26x - 10 = 0\) is:
\[ \boxed{-\frac{2}{3}} \]