Let α,β,γ be the roots of x^3-2x^2+3x-5=0. Which of the following is/are correct?

From Vieta, $e_1=2$, $e_2=3$, $e_3=5$. Then $\sum\alpha^2=e_1^2-2e_2=4-6=-2$, so (A) is true, and the negative value just reflects the complex roots. By Newton's identity, $\sum\alpha^3=e_1\sum\alpha^2-e_2e_1+3e_3=-4-6+15=5$, so (B) with $9$ is false. The discriminant of the cubic is $-367<0$, so there is one real root and a complex pair; (C) is true. The new cubic has coefficients: sum of roots $=-2$ giving $x^2$ coefficient $+2$; sum of products $=e_2^2-2e_1e_3=9-20=-11$; product $=e_3^2=25$ giving constant $-25$. Hence $x^3+2x^2-11x-25=0$, and (D) is true.