<s>[INST] Find $k$ in $p(x) = 2x^3 - 6x^2 + kx -1$ such that its roots $x_1^2+x_2^2+x_3^2 = 6$ Let
$$p(x) = 2x^3 - 6x^2 + kx -1$$
and let $x_1, x_2$
and $x_3$ the $p(x)$ roots. What is the $k$ value such that
$$x_1^2+x_2^2+x_3^2 = 6$$
 [/INST] Using Newton’s identities,
$$x_1^2+x_2^2+x_3^2= (x_1+x_2+x_3)^2- 2\sum x_ix_j=3^2-k=6$$
and therefore $k=3$.
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