Construct a quadratic Equation whose roots are; (a) \( \alpha+\frac{1}{\beta}, \beta+\frac{1}{\alpha} \) (b) \( \alpha^{3} \) and \( \beta^{3} \)

Let \[ s=\alpha+\beta \quad\text{and}\quad p=\alpha\beta. \] ### (a) Roots: \(\alpha+\frac{1}{\beta}\) and \(\beta+\frac{1}{\alpha}\) **Step 1. Sum of the roots** We have: \[ \left(\alpha+\frac{1}{\beta}\right)+\left(\beta+\frac{1}{\alpha}\right)=(\alpha+\beta)+\left(\frac{1}{\alpha}+\frac{1}{\beta}\right). \] Since \[ \frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{s}{p}, \] the sum becomes: \[ s+\frac{s}{p}=\frac{s(p+1)}{p}. \] **Step 2. Product of the roots** Compute: \[ \left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right). \] Expanding, \[ \left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right) =\alpha\beta+\alpha\left(\frac{1}{\alpha}\right)+\frac{1}{\beta}\cdot\beta+\frac{1}{\alpha\beta} =\alpha\beta+1+1+\frac{1}{\alpha\beta}. \] That is, \[ p+2+\frac{1}{p}=\frac{p^2+2p+1}{p}=\frac{(p+1)^2}{p}. \] **Step 3. Constructing the Equation** If a quadratic equation has roots \(r_1\) and \(r_2\), then it can be written as: \[ x^2 - (r_1+r_2)x + r_1r_2=0. \] Substitute \(r_1+r_2=\frac{s(p+1)}{p}\) and \(r_1r_2=\frac{(p+1)^2}{p}\): \[ x^2 - \frac{s(p+1)}{p}x + \frac{(p+1)^2}{p}=0. \] Multiplying through by \(p\) to eliminate the denominator gives: \[ p\,x^2 - s(p+1)x + (p+1)^2=0. \] ### (b) Roots: \(\alpha^3\) and \(\beta^3\) **Step 1. Sum of the roots** We have: \[ \alpha^3+\beta^3. \] Using the identity: \[ \alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta), \] this becomes: \[ s^3-3ps. \] **Step 2. Product of the roots** Since: \[ \alpha^3\beta^3=(\alpha\beta)^3=p^3, \] the product is \(p^3\). **Step 3. Constructing the Equation** The quadratic equation is: \[ x^2 - (\,\text{sum of roots}\,)x + (\text{product of roots})=0. \] Thus, \[ x^2 - (s^3-3ps)x + p^3=0. \] ### Final Answers (a) The required quadratic equation is: \[ p\,x^2 - s(p+1)x + (p+1)^2=0. \] (b) The required quadratic equation is: \[ x^2 - (s^3-3ps)x + p^3=0. \]