If `alpha` and`beta` are zeroes of a polynomial f(x0`=3x^(2)-4x+1`, finda quadratic polynomial whose zeroes are `(alpha^(2))/(beta)` and `(beta^(2))/(alpha)`.

Explanation

To find a quadratic polynomial whose zeroes are \(\frac{\alpha^2}{\beta}\) and \(\frac{\beta^2}{\alpha}\), we will follow these steps:

### Step 1: Identify the given polynomial and its roots
The polynomial given is:
\[ f(x) = 3x^2 - 4x + 1 \]
The roots of this polynomial are denoted as \(\alpha\) and \(\beta\).

### Step 2: Calculate the sum and product of the roots
Using Vieta's formulas:
- The sum of the roots \(\alpha + \beta = -\frac{b}{a} = \frac{4}{3}\)
- The product of the roots \(\alpha \beta = \frac{c}{a} = \frac{1}{3}\)

### Step 3: Find the new roots
We need to find the new roots:
1. \(\frac{\alpha^2}{\beta}\)
2. \(\frac{\beta^2}{\alpha}\)

#### Step 3.1: Calculate the sum of the new roots
The sum of the new roots is:
\[ \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} \]
This can be simplified as follows:
\[ = \frac{\alpha^3 + \beta^3}{\alpha \beta} \]
Using the identity for the sum of cubes:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) \]
We can express \(\alpha^2 + \beta^2\) using:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Substituting the values:
\[ \alpha^2 + \beta^2 = \left(\frac{4}{3}\right)^2 - 2 \cdot \frac{1}{3} = \frac{16}{9} - \frac{2}{3} = \frac{16}{9} - \frac{6}{9} = \frac{10}{9} \]
Now substituting back:
\[ \alpha^3 + \beta^3 = \left(\frac{4}{3}\right)\left(\frac{10}{9} - \frac{1}{3}\right) = \left(\frac{4}{3}\right)\left(\frac{10}{9} - \frac{3}{9}\right) = \left(\frac{4}{3}\right)\left(\frac{7}{9}\right) = \frac{28}{27} \]
Thus, the sum of the new roots is:
\[ \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} = \frac{\frac{28}{27}}{\frac{1}{3}} = \frac{28}{27} \cdot 3 = \frac{84}{27} = \frac{28}{9} \]

#### Step 3.2: Calculate the product of the new roots
The product of the new roots is:
\[ \frac{\alpha^2}{\beta} \cdot \frac{\beta^2}{\alpha} = \frac{\alpha^2 \beta^2}{\alpha \beta} = \alpha \beta \]
Thus, the product of the new roots is:
\[ \alpha \beta = \frac{1}{3} \]

### Step 4: Form the quadratic polynomial
The quadratic polynomial with roots \(\frac{\alpha^2}{\beta}\) and \(\frac{\beta^2}{\alpha}\) can be expressed as:
\[ x^2 - \text{(sum of roots)} \cdot x + \text{(product of roots)} \]
Substituting the values we found:
\[ x^2 - \frac{28}{9}x + \frac{1}{3} \]

### Step 5: Clear the fractions
To eliminate the fractions, multiply the entire polynomial by 9:
\[ 9x^2 - 28x + 3 \]

### Final Answer
The required quadratic polynomial is:
\[ 9x^2 - 28x + 3 \]