If one real root of the quadratic equation `81x^2 + kx + 256=0` is cube of the other root, then a value of k is

Options

A. `-300`
B. `-81`
C. 144
D. 100

Explanation

To solve the problem, we need to find the value of \( k \) in the quadratic equation \( 81x^2 + kx + 256 = 0 \) given that one root is the cube of the other root.

Let's denote the roots of the quadratic equation as \( \alpha \) and \( \beta \). According to the problem, we can assume that \( \beta = \alpha^3 \).

### Step 1: Use the relationships of the roots
From Vieta's formulas, we know:
- The sum of the roots \( \alpha + \beta = -\frac{k}{81} \)
- The product of the roots \( \alpha \cdot \beta = \frac{256}{81} \)

Substituting \( \beta = \alpha^3 \) into these equations, we get:
1. \( \alpha + \alpha^3 = -\frac{k}{81} \) (Equation 1)
2. \( \alpha \cdot \alpha^3 = \frac{256}{81} \) (Equation 2)

### Step 2: Simplify the product of the roots
From Equation 2:
\[ \alpha^4 = \frac{256}{81} \]
Taking the fourth root of both sides, we find:
\[ \alpha = \pm \sqrt[4]{\frac{256}{81}} = \pm \frac{4}{3} \]

### Step 3: Substitute \( \alpha \) back into the sum of the roots
Now we will consider both cases for \( \alpha \).

**Case 1:** \( \alpha = \frac{4}{3} \)

Substituting this value into Equation 1:
\[ \frac{4}{3} + \left(\frac{4}{3}\right)^3 = -\frac{k}{81} \]
Calculating \( \left(\frac{4}{3}\right)^3 \):
\[ \left(\frac{4}{3}\right)^3 = \frac{64}{27} \]
Now substituting this back:
\[ \frac{4}{3} + \frac{64}{27} = -\frac{k}{81} \]
To add these fractions, we need a common denominator:
\[ \frac{4}{3} = \frac{36}{27} \]
Thus:
\[ \frac{36}{27} + \frac{64}{27} = \frac{100}{27} \]
So we have:
\[ \frac{100}{27} = -\frac{k}{81} \]

### Step 4: Cross-multiply to solve for \( k \)
Cross-multiplying gives:
\[ 100 \cdot 81 = -27k \]
Calculating \( 100 \cdot 81 = 8100 \):
\[ 8100 = -27k \]
Thus:
\[ k = -\frac{8100}{27} = -300 \]

**Case 2:** \( \alpha = -\frac{4}{3} \)

Substituting this value into Equation 1:
\[ -\frac{4}{3} + \left(-\frac{4}{3}\right)^3 = -\frac{k}{81} \]
Calculating \( \left(-\frac{4}{3}\right)^3 \):
\[ \left(-\frac{4}{3}\right)^3 = -\frac{64}{27} \]
Now substituting this back:
\[ -\frac{4}{3} - \frac{64}{27} = -\frac{k}{81} \]
Again, converting \( -\frac{4}{3} \) to have a common denominator:
\[ -\frac{4}{3} = -\frac{36}{27} \]
Thus:
\[ -\frac{36}{27} - \frac{64}{27} = -\frac{100}{27} \]
So we have:
\[ -\frac{100}{27} = -\frac{k}{81} \]
Cross-multiplying gives:
\[ 100 \cdot 81 = 27k \]
Thus:
\[ k = \frac{8100}{27} = 300 \]

### Conclusion
The values of \( k \) we found are \( -300 \) and \( 300 \). Since the problem asks for a value of \( k \), we can conclude that one possible value of \( k \) is:
\[ \boxed{-300} \]