Consider the biquadratic equation $x^4+4px^3+6qx^2+4rx+s=0$. If $\alpha_{1}, \alpha_{2}, \alpha_{3}$ and $\alpha_{4}$ are the roots of this equation find— $\sum_{\text {symmetric}}\left(\alpha_{1}-\alpha_{2}\right)^{2}$ and $\sum_{\text {symmetric}}\left(\alpha_{1}-\alpha_{2}\right)^{4}$.
What is your answer for $\sum_{\text {symmetric }}\left(\alpha_{1}-\alpha_{2}\right)^{2 k}$ where $k$ is a natural number. Find also $\left(\alpha_{1}+\alpha_{2}-\alpha_{3}-\alpha_{4}\right)\left(\alpha_{1}+\alpha_{3}-\alpha_{2}-\alpha_{4}\right)\left(\alpha_{1}+\alpha_{4}-\alpha_{2}-\alpha_{3}\right)$.
To be honest, I have no idea how to solve the last section, but anyways I think that an elegant application of Newton's identities should provide us with the solution.
I am able to solve for $\sum_{\text {symmetric}}\left(\alpha_{1}-\alpha_{2}\right)^{2}$ which comes out to be $96(p^2-q)$ by using Vieta's relations. But doing the same for $\sum_{\text {symmetric}}\left(\alpha_{1}-\alpha_{2}\right)^{4}$ makes it too complex.
I'm looking for an elegant solution or atleast a hint for the above problems.
Any help would be appreciated.