If `f(x)=64x^(3)+(1)/(x^(3))` and `alpha, beta` are the roots of `4x+(1)/(x)=2` then :

Options

A. `f(alpha)=-24`
B. `f(alpha)=-16`
C. `f(beta)=-16`
D. both (b) and (c )

Answer

To solve the problem, we need to find the value of \( f(\alpha) \) and \( f(\beta) \) given the function \( f(x) = 64x^3 + \frac{1}{x^3} \) and the roots \( \alpha \) and \( \beta \) of the equation \( 4x + \frac{1}{x} = 2 \).

### Step-by-Step Solution:

1. **Identify the equation and its roots**:
The equation given is:
\[ 4x + \frac{1}{x} = 2 \]
Let's denote this as Equation (1).

2. **Rearranging Equation (1)**:
We can rearrange Equation (1) to isolate the term involving \( x \):
\[ 4x = 2 - \frac{1}{x} \]
Multiplying through by \( x \) (assuming \( x \neq 0 \)):
\[ 4x^2 = 2x - 1 \]
Rearranging gives us:
\[ 4x^2 - 2x + 1 = 0 \]

3. **Finding the roots of the quadratic**:
We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 4, b = -2, c = 1 \):
\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 4 \cdot 1}}{2 \cdot 4} \]
\[ x = \frac{2 \pm \sqrt{4 - 16}}{8} \]
\[ x = \frac{2 \pm \sqrt{-12}}{8} \]
\[ x = \frac{2 \pm 2i\sqrt{3}}{8} = \frac{1 \pm i\sqrt{3}}{4} \]
Thus, the roots \( \alpha \) and \( \beta \) are:
\[ \alpha = \frac{1 + i\sqrt{3}}{4}, \quad \beta = \frac{1 - i\sqrt{3}}{4} \]

4. **Cubing both sides of Equation (1)**:
We cube both sides of the original equation:
\[ (4x + \frac{1}{x})^3 = 2^3 \]
Expanding the left side using the binomial theorem:
\[ 64x^3 + 3 \cdot 4x \cdot 4x \cdot \frac{1}{x} + 3 \cdot 4x \cdot \frac{1}{x^2} + \frac{1}{x^3} = 8 \]
Simplifying gives:
\[ 64x^3 + 12 + \frac{1}{x^3} = 8 \]

5. **Rearranging to find \( f(x) \)**:
Now, we can express \( f(x) \):
\[ 64x^3 + \frac{1}{x^3} = 8 - 12 = -4 \]
Thus, we have:
\[ f(x) = -4 \]

6. **Conclusion**:
Since \( \alpha \) and \( \beta \) are roots of the equation, we have:
\[ f(\alpha) = f(\beta) = -4 \]

### Final Answer:
The values of \( f(\alpha) \) and \( f(\beta) \) are both equal to \( -4 \).