The equation \( 3 x^{2}+3 k x+k=0 \) where \( k \) is ion-zero constant, has roots \( \alpha \) and \( \beta \). find the quadratic equation whose oots are \( \frac{\alpha^{2}}{\beta\left(\alpha^{3}+\beta^{3}+k^{3}\right)} \) and \( \frac{\beta^{2}}{\alpha\left(\alpha^{3}+\beta^{3}+k^{3}\right)} \)

To find the quadratic equation whose roots are given as \( \frac{\alpha^{2}}{\beta(\alpha^{3}+\beta^{3}+k^{3})} \) and \( \frac{\beta^{2}}{\alpha(\alpha^{3}+\beta^{3}+k^{3})} \), we will first need to calculate the sum and product of the roots. Using Vieta's formulas on the original quadratic, the sum \( S \) and product \( P \) of roots \( \alpha \) and \( \beta \) can be derived as follows: 1. The sum of the roots \( S = \alpha + \beta = -\frac{3k}{3} = -k \). 2. The product of the roots \( P = \alpha \beta = \frac{k}{3} \). Next, we calculate the new required sum and product: - The new sum of roots: \[ S' = \frac{\alpha^{2}}{\beta(\alpha^{3}+\beta^{3}+k^{3})} + \frac{\beta^{2}}{\alpha(\alpha^{3}+\beta^{3}+k^{3})} = \frac{\alpha^3 + \beta^3}{\alpha\beta(\alpha^{3}+\beta^{3}+k^{3})} \] Using the identity \( \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) = S(S^2 - 3P) \): \[ \alpha^3 + \beta^3 = -k((-k)^2 - 3\cdot\frac{k}{3}) = -k(k^2 - k) = -k^3 + k^2 \] Putting this into \( S' \): \[ S' = \frac{-k^3 + k^2}{\frac{k}{3}(\alpha^{3}+\beta^{3}+k^{3})} = \frac{-k(k - 1)}{(\alpha^{3}+\beta^{3}+k^{3})/3} \] - The new product of roots: \[ P' = \frac{\alpha^2\beta^2}{\alpha\beta(\alpha^{3}+\beta^{3}+k^{3})} = \frac{P^2}{\alpha^{3}+\beta^{3}+k^{3}} = \frac{\left( \frac{k}{3} \right)^{2}}{\alpha^{3}+\beta^{3}+k^{3}} \] Now, using the new sums \( S' \) and products \( P' \), the quadratic equation can be written in the form \( x^2 - S'x + P' = 0 \). Thus, the final quadratic equation will take the following form after substituting our values for \( S' \) and \( P' \). The process can be long, but with a consistent analysis of roots, you can generate the necessary equation accordingly!