If $\alpha,\beta$ are the roots of $\displaystyle x^{2}-px+q=0$, then the product of the roots of the quadratic equation whose roots are $\displaystyle \alpha^{2}-\beta^{2}$ and $\alpha^{3}-\beta^{3}$ is
none of these
Roots are alpha, beta. alpha+beta = p, alpha*beta = q. New roots: alpha^2-beta^2 = (alpha-beta)(alpha+beta) and alpha^3-beta^3 = (alpha-beta)(alpha^2+alpha*beta+beta^2). Product = (alpha-beta)^2 * (alpha+beta) * (alpha^2+alpha*beta+beta^2). (alpha-beta)^2 = (alpha+beta)^2 - 4alpha*beta = p^2-4q. alpha^2+alpha*beta+beta^2 = (alpha+beta)^2 - alpha*beta = p^2-q. Product = (p^2-4q) * p * (p^2-q).
From the original equation, the sum of the roots p equals alpha plus beta and the product q equals alpha beta. The product of the roots of the new equation is (alpha squared minus beta squared) times (alpha cubed minus beta cubed), which factors to (alpha minus beta)(alpha plus beta) times (alpha minus beta)(alpha squared plus alpha beta plus beta squared). Substituting p and q, this becomes p times (alpha minus beta) squared times (p squared minus q). We find (alpha minus beta) squared equals (alpha plus beta) squared minus 4 times alpha beta, which is p squared minus 4q. Replacing this in the expression gives the final product as p times (p squared minus 4q) times (p squared minus q). The result is p(p squared minus q)(p squared minus 4q).