If `alpha,beta` are the roots of the equation `8x^2-3x+27=0,` then the value of `(alpha^2/beta)^(1/3)+(beta^2/alpha)^(1/3)` is

Options

A. `1/3`
B. `1/4`
C. `1/5`
D. `1/6`

Answer

To solve the problem, we need to find the value of \((\frac{\alpha^2}{\beta})^{1/3} + (\frac{\beta^2}{\alpha})^{1/3}\) given that \(\alpha\) and \(\beta\) are the roots of the equation \(8x^2 - 3x + 27 = 0\).

### Step 1: Identify the coefficients
The given quadratic equation is \(8x^2 - 3x + 27 = 0\). Here, we can identify:
- \(a = 8\)
- \(b = -3\)
- \(c = 27\)

### Step 2: Use Vieta's formulas
From Vieta's formulas, we know:
- The sum of the roots \(\alpha + \beta = -\frac{b}{a} = -\frac{-3}{8} = \frac{3}{8}\)
- The product of the roots \(\alpha \beta = \frac{c}{a} = \frac{27}{8}\)

### Step 3: Substitute into the expression
We need to evaluate the expression:
\[ (\frac{\alpha^2}{\beta})^{1/3} + (\frac{\beta^2}{\alpha})^{1/3} \]

This can be rewritten as:
\[ \frac{\alpha^{2/3}}{\beta^{1/3}} + \frac{\beta^{2/3}}{\alpha^{1/3}} \]

### Step 4: Combine the terms
To combine the terms, we can take a common denominator:
\[ \frac{\alpha^{2/3} \cdot \alpha^{1/3} + \beta^{2/3} \cdot \beta^{1/3}}{\alpha^{1/3} \cdot \beta^{1/3}} = \frac{\alpha^{1} + \beta^{1}}{\alpha^{1/3} \cdot \beta^{1/3}} \]
This simplifies to:
\[ \frac{\alpha + \beta}{(\alpha \beta)^{1/3}} \]

### Step 5: Substitute values from Vieta's
Substituting the values we found from Vieta's:
\[ \alpha + \beta = \frac{3}{8}, \quad \alpha \beta = \frac{27}{8} \]

Now substituting these values into the expression:
\[ \frac{\frac{3}{8}}{(\frac{27}{8})^{1/3}} \]

### Step 6: Simplify the denominator
Calculating the cube root:
\[ (\frac{27}{8})^{1/3} = \frac{3}{2} \]

### Step 7: Final calculation
Now substituting back:
\[ \frac{\frac{3}{8}}{\frac{3}{2}} = \frac{3}{8} \cdot \frac{2}{3} = \frac{2}{8} = \frac{1}{4} \]

### Conclusion
Thus, the value of \((\frac{\alpha^2}{\beta})^{1/3} + (\frac{\beta^2}{\alpha})^{1/3}\) is \(\frac{1}{4}\).