If one root is cube of the other of equation `81x^2 +kx+256=0` then value of k is equal to (A) `100` (B) `-300` (C) `-81` (D) `400`

Options

A. 100
B. -300
C. -81
D. 400

Answer

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 as \( \alpha \) and \( \alpha^3 \).

### Step 1: Use Vieta's Formulas
According to Vieta's formulas, for a quadratic equation \( ax^2 + bx + c = 0 \):
- The sum of the roots \( \alpha + \alpha^3 = -\frac{k}{81} \)
- The product of the roots \( \alpha \cdot \alpha^3 = \frac{256}{81} \)

### Step 2: Express the Product of Roots
From the product of the roots, we have:
\[ \alpha \cdot \alpha^3 = \alpha^4 = \frac{256}{81} \]
This implies:
\[ \alpha^4 = \frac{256}{81} \]

### Step 3: Solve for \( \alpha \)
Taking the fourth root of both sides gives us:
\[ \alpha = \sqrt[4]{\frac{256}{81}} = \frac{\sqrt[4]{256}}{\sqrt[4]{81}} = \frac{4}{3} \]
Thus, \( \alpha = \frac{4}{3} \).

### Step 4: Find the Sum of the Roots
Now substituting \( \alpha \) back into the sum of the roots:
\[ \alpha + \alpha^3 = \frac{4}{3} + \left(\frac{4}{3}\right)^3 \]
Calculating \( \left(\frac{4}{3}\right)^3 \):
\[ \left(\frac{4}{3}\right)^3 = \frac{64}{27} \]
Now, we need a common denominator to add \( \frac{4}{3} \) and \( \frac{64}{27} \):
\[ \frac{4}{3} = \frac{36}{27} \]
Thus,
\[ \alpha + \alpha^3 = \frac{36}{27} + \frac{64}{27} = \frac{100}{27} \]

### Step 5: Substitute into Vieta's Formula
Now substituting into Vieta's formula for the sum of the roots:
\[ \frac{100}{27} = -\frac{k}{81} \]
Cross-multiplying gives:
\[ 100 \cdot 81 = -27k \]
\[ 8100 = -27k \]
\[ k = -\frac{8100}{27} = -300 \]

### Conclusion
Thus, the value of \( k \) is \( -300 \).

### Final Answer
The value of \( k \) is \( \boxed{-300} \).