2024 SSMO Team Round Problems/Problem 6

Problem

Let $\alpha$ , $\beta$ , and $\gamma$ be the roots of the polynomial $x^3 - 6x^2 - 19x - n$ . If $n$ is an integer, what is the least possible positive value of $\alpha^3 + \beta^3 + \gamma^3$ ?

Solution

From Vieta's Formulas, we have $\alpha+\beta+\gamma = 6,\alpha\beta+\alpha\gamma+\beta\gamma = -19,\text{ and }\alpha\beta\gamma = n.$ Since $x^3-6x^2-19x-n = 0$ for $x \in \{\alpha,\beta,\gamma\},$ we have $x^3 = 6x^2+19x+n.$ So, \begin{align*} \alpha^3+\beta^3+\gamma^3&=(6\alpha^2+19\alpha+n)+(6\beta^2+19\beta+n)\\&+(6\gamma^2+19\gamma+n)\\ &=6(\alpha^2+\beta^2+\gamma^2)+19(\alpha+\beta+\gamma)+3n\\ &=6((\alpha+\beta+\gamma)^2-2(\alpha\beta+\alpha\gamma+\beta\gamma))\\ &+19(\alpha+\beta+\gamma)+3n\\ &=6((6^2)-2(-19))+19(6)+3n = 558+3n. \end{align*} Since $n$ is an integer and we are seeking to find the least positive value of $558+3n = 3(n+186),$ we let $n = -185,$ giving an answer of $\boxed{3}.$

~SMO_Team

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2024 SSMO Team Round Problems/Problem 6

Problem

Solution