How to derive this formula involving the characteristic polynomial of a square matrix of order $3$?

Question

Let $P_A$ denote the characteristic polynomial for a $3\times3$ matrix $A$. Then $\newcommand{\Tr}{\text{Tr}}$ $$ P_A(x)=\frac{1}{6}[\Tr^3(A)+2\Tr(A^3)-3\Tr(A)\Tr(A^2)]-\frac{1}{2}[\Tr^2(A)-\Tr(A^2)]x+\Tr(A)x^2-x^3.$$ I am wondering if there is a way to derive this without naively calculating each term and comparing whether this equal to the determinant $\det(A-xI).$

Answer 1

This is an application of Newton's identities; the coefficients of the characteristic polynomial are elementary symmetric polynomials in the eigenvalues of $A$, and the traces of the powers of $A$ are power sums of the eigenvalues of $A$.