If `alpha` and `beta` are zeroes of `8x^(2)-6x+1`, then find the value of `(1)/(alpha)+(1)/(beta)`.

Explanation

To find the value of \( \frac{1}{\alpha} + \frac{1}{\beta} \) where \( \alpha \) and \( \beta \) are the zeros of the polynomial \( 8x^2 - 6x + 1 \), we can follow these steps:

### Step 1: Identify coefficients
The given polynomial is \( 8x^2 - 6x + 1 \). Here, we can identify the coefficients:
- \( a = 8 \)
- \( b = -6 \)
- \( c = 1 \)

### Step 2: Calculate the sum of the zeros
The sum of the zeros \( \alpha + \beta \) can be calculated using the formula:
\[ \alpha + \beta = -\frac{b}{a} \]
Substituting the values of \( b \) and \( a \):
\[ \alpha + \beta = -\frac{-6}{8} = \frac{6}{8} = \frac{3}{4} \]

### Step 3: Calculate the product of the zeros
The product of the zeros \( \alpha \beta \) can be calculated using the formula:
\[ \alpha \beta = \frac{c}{a} \]
Substituting the values of \( c \) and \( a \):
\[ \alpha \beta = \frac{1}{8} \]

### Step 4: Calculate \( \frac{1}{\alpha} + \frac{1}{\beta} \)
Using the relationship:
\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} \]
Substituting the values we found:
\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\frac{3}{4}}{\frac{1}{8}} \]

### Step 5: Simplify the expression
To simplify \( \frac{\frac{3}{4}}{\frac{1}{8}} \), we multiply by the reciprocal:
\[ \frac{3}{4} \times 8 = \frac{3 \times 8}{4} = \frac{24}{4} = 6 \]

### Conclusion
Thus, the value of \( \frac{1}{\alpha} + \frac{1}{\beta} \) is \( 6 \).

---