If roots of quadrratic equation `x^2-64x+256=0` are `alpha` & `beta` then `(alpha^3/beta^5)^(1/8)+(beta^3/alpha^5)^(1/8)` =

Options

A. `2`
B. `6`
C. `-2`
D. `5`

Answer

To solve the problem, we start with the quadratic equation given:

\[ x^2 - 64x + 256 = 0 \]

### Step 1: Find the roots \( \alpha \) and \( \beta \)

Using Vieta's formulas, we know:
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} = 64 \)
- The product of the roots \( \alpha \beta = \frac{c}{a} = 256 \)

### Step 2: Calculate \( \frac{\alpha^3}{\beta^5} \) and \( \frac{\beta^3}{\alpha^5} \)

We need to evaluate:

\[ \left( \frac{\alpha^3}{\beta^5} \right)^{\frac{1}{8}} + \left( \frac{\beta^3}{\alpha^5} \right)^{\frac{1}{8}} \]

### Step 3: Combine the terms under a common base

We can rewrite the expression as:

\[ \left( \frac{\alpha^3}{\beta^5} \right)^{\frac{1}{8}} + \left( \frac{\beta^3}{\alpha^5} \right)^{\frac{1}{8}} = \frac{\alpha^{\frac{3}{8}}}{\beta^{\frac{5}{8}}} + \frac{\beta^{\frac{3}{8}}}{\alpha^{\frac{5}{8}}} \]

### Step 4: Find a common denominator

The common denominator for the two fractions is \( \beta^{\frac{5}{8}} \alpha^{\frac{5}{8}} \):

\[ = \frac{\alpha^{\frac{3}{8}} \alpha^{\frac{5}{8}} + \beta^{\frac{3}{8}} \beta^{\frac{5}{8}}}{\beta^{\frac{5}{8}} \alpha^{\frac{5}{8}}} \]

This simplifies to:

\[ = \frac{\alpha^{\frac{8}{8}} + \beta^{\frac{8}{8}}}{\beta^{\frac{5}{8}} \alpha^{\frac{5}{8}}} = \frac{\alpha + \beta}{\beta^{\frac{5}{8}} \alpha^{\frac{5}{8}}} \]

### Step 5: Substitute known values

We know from Step 1 that \( \alpha + \beta = 64 \) and \( \alpha \beta = 256 \). Therefore, we can express \( \beta^{\frac{5}{8}} \alpha^{\frac{5}{8}} \) as:

\[ (\alpha \beta)^{\frac{5}{8}} = 256^{\frac{5}{8}} = (2^8)^{\frac{5}{8}} = 2^{5} = 32 \]

### Step 6: Final calculation

Now substituting back into our expression:

\[ = \frac{64}{32} = 2 \]

Thus, the final answer is:

\[ \boxed{2} \]