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 start with the quadratic equation given:

\[ 81x^2 + kx + 256 = 0 \]

We are told that one root is the cube of the other root. Let's denote the roots as \( \alpha \) and \( \alpha^3 \).

### Step 1: Use the product of the roots
According to Vieta's formulas, the product of the roots of a quadratic equation \( ax^2 + bx + c = 0 \) is given by:

\[ \text{Product of roots} = \frac{c}{a} \]

For our equation:

\[ \alpha \cdot \alpha^3 = \frac{256}{81} \]

This simplifies to:

\[ \alpha^4 = \frac{256}{81} \]

### Step 2: Solve for \( \alpha \)
Now, we can express \( \alpha \) in terms of its fourth root:

\[ \alpha = \pm \sqrt[4]{\frac{256}{81}} \]

Calculating the fourth root:

\[ \alpha = \pm \frac{\sqrt{256}}{\sqrt{81}} = \pm \frac{16}{9} \]

### Step 3: Use the sum of the roots
Next, we use the sum of the roots, which is also given by Vieta's formulas:

\[ \text{Sum of roots} = -\frac{b}{a} = -\frac{k}{81} \]

Thus, we have:

\[ \alpha + \alpha^3 = -\frac{k}{81} \]

### Step 4: Calculate \( \alpha + \alpha^3 \)
Now we will calculate \( \alpha + \alpha^3 \) for both cases of \( \alpha \).

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

\[ \alpha^3 = \left(\frac{4}{3}\right)^3 = \frac{64}{27} \]

Therefore,

\[ \alpha + \alpha^3 = \frac{4}{3} + \frac{64}{27} \]

To add these fractions, we need a common denominator:

\[ \frac{4}{3} = \frac{36}{27} \]

So,

\[ \alpha + \alpha^3 = \frac{36}{27} + \frac{64}{27} = \frac{100}{27} \]

Setting this equal to \( -\frac{k}{81} \):

\[ \frac{100}{27} = -\frac{k}{81} \]

Cross-multiplying gives:

\[ 100 \cdot 81 = -27k \implies 8100 = -27k \implies k = -\frac{8100}{27} = -300 \]

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

\[ \alpha^3 = \left(-\frac{4}{3}\right)^3 = -\frac{64}{27} \]

Therefore,

\[ \alpha + \alpha^3 = -\frac{4}{3} - \frac{64}{27} \]

Again, using a common denominator:

\[ -\frac{4}{3} = -\frac{36}{27} \]

So,

\[ \alpha + \alpha^3 = -\frac{36}{27} - \frac{64}{27} = -\frac{100}{27} \]

Setting this equal to \( -\frac{k}{81} \):

\[ -\frac{100}{27} = -\frac{k}{81} \]

Cross-multiplying gives:

\[ 100 \cdot 81 = 27k \implies 8100 = 27k \implies k = \frac{8100}{27} = 300 \]

### Conclusion
Thus, we have two possible values for \( k \):
- When \( \alpha = \frac{4}{3} \), \( k = -300 \)
- When \( \alpha = -\frac{4}{3} \), \( k = 300 \)

Since the problem specifies that we need a value of \( k \), and the options provided only include \( -300 \), we conclude:

**Final Answer: \( k = -300 \)**