If `f (x) = 27x^(3) -(1)/(x^(3))` and `alpha, beta` are roots of `3x - (1)/(x) = 2` then

Options

A. `f (alpha) = f(beta)`
B. `f(alpha) =10`
C. `f (beta) = -10`
D. none of these

Explanation

To solve the problem step by step, we start with the given function and the roots of the equation.

### Step 1: Identify the function and the equation
We have:
\[ f(x) = 27x^3 - \frac{1}{x^3} \]
and the roots \( \alpha \) and \( \beta \) satisfy the equation:
\[ 3x - \frac{1}{x} = 2 \]

### Step 2: Solve for \( \alpha \) and \( \beta \)
To find \( \alpha \) and \( \beta \), we rearrange the equation:
\[ 3x - \frac{1}{x} = 2 \]
Multiplying through by \( x \) (assuming \( x \neq 0 \)):
\[ 3x^2 - 1 = 2x \]
Rearranging gives us:
\[ 3x^2 - 2x - 1 = 0 \]

### Step 3: Use the quadratic formula
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3, b = -2, c = -1 \):
\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} \]
\[ x = \frac{2 \pm \sqrt{4 + 12}}{6} \]
\[ x = \frac{2 \pm \sqrt{16}}{6} \]
\[ x = \frac{2 \pm 4}{6} \]

Calculating the two roots:
1. \( x = \frac{6}{6} = 1 \)
2. \( x = \frac{-2}{6} = -\frac{1}{3} \)

Thus, \( \alpha = 1 \) and \( \beta = -\frac{1}{3} \).

### Step 4: Calculate \( f(\alpha) \) and \( f(\beta) \)
Now we calculate \( f(\alpha) \) and \( f(\beta) \):
1. For \( \alpha = 1 \):
\[ f(1) = 27(1)^3 - \frac{1}{(1)^3} = 27 - 1 = 26 \]

2. For \( \beta = -\frac{1}{3} \):
\[ f\left(-\frac{1}{3}\right) = 27\left(-\frac{1}{3}\right)^3 - \frac{1}{\left(-\frac{1}{3}\right)^3} \]
\[ = 27\left(-\frac{1}{27}\right) - \frac{1}{-\frac{1}{27}} \]
\[ = -1 + 27 = 26 \]

### Step 5: Conclusion
Thus, we find:
\[ f(\alpha) = 26 \]
\[ f(\beta) = 26 \]

Both roots give the same function value.