AMC 12 Mock Exam 1 Problem 15

Let $e_1=x+y+z=6$, $e_2=xy+yz+zx$, $e_3=xyz$, and let $p_k=x^k+y^k+z^k$. From $p_2=e_1^2-2e_2$ we get $14=36-2e_2$, so $e_2=11$. From $p_3=e_1^3-3e_1e_2+3e_3$ we get $36=216-198+3e_3$, so $e_3=6$.

Each of $x,y,z$ satisfies $t^3=e_1t^2-e_2t+e_3$. Multiplying by $t$ and summing over $t=x,y,z$ gives $p_4=e_1p_3-e_2p_2+e_3p_1=6\cdot36-11\cdot14+6\cdot6=216-154+36=\boxed{98}$.

(Indeed $x,y,z$ are the roots of $t^3-6t^2+11t-6=(t-1)(t-2)(t-3)$, and $1+16+81=98$.)