What is $\mbox{Tr}^2(A)-\mbox{Tr}(A^2)$ in terms of the eigenvalues of $A$?

I am looking for a way to relate the terms of the characteristic polynomial of a $3 \times 3$ matrix to its eigenvalues.

$\\P_{3}(A)=x^{3} + \mbox{Tr}(A)x^{2} + (\mbox{Tr}^{2}(A) - \mbox{Tr}(A^{2}))x^1 +(\mbox{Tr}^{3}(A) + 2 \mbox{Tr} (A^{3}) -3 \mbox{Tr} (A) \mbox{Tr}(A^{2}))x^0$

Which can also be written as

$\\P_{3}(A)=x^{3} + \mbox{Tr}(A)x^{2} + (\mbox{Tr}^{2}(A) - \mbox{Tr}(A^{2}))x^1 + \det(A)x^0$,

which again can be rewritten using the properties of eigenvalues as

$\\P_{3}(A)=x^{3} + x^{2} \sum_{i=1}^{i=3} \lambda_{i} + x^1(\mbox{Tr}^{2}(A) - \mbox{Tr}(A^{2})) + x^0 \prod_{i=1}^{i=3} \lambda_{i}$ .

I am wondering is there a way to relate the term in $x$ also to the eigenvalues in a similar way that can be done in with the terms in $x^{2}$ and $x^{0}$?

Thanks!

$$\operatorname{Tr}^2(A)-\operatorname{Tr}(A^2)=(\lambda_1+\lambda_2+\lambda_3)^2-(\lambda_1^2+\lambda_2^2+\lambda_3^2)=2(\lambda_1\lambda_2+\lambda_1\lambda_3+\lambda_1\lambda_3)$$