For the cubic polynomial $p(x)=x^3-6x^2+11x-6$, let its zeros be $\alpha ,\beta ,\gamma$. Find ${\alpha }^2+{\beta }^2+{\gamma }^2$.

Step-by-step solution

1. From Vieta’s formulas: $\alpha +\beta +\gamma =6$, $\alpha \beta +\beta \gamma +\gamma \alpha =11$.
2. Recall ${\alpha }^2+{\beta }^2+{\gamma }^2=(\alpha +\beta +\gamma {)}^2-2(\alpha \beta +\beta \gamma +\gamma \alpha )$.
3. Compute $(\alpha +\beta +\gamma {)}^2=6^2=36$.
4. Compute $2(\alpha \beta +\beta \gamma +\gamma \alpha )=2\times 11=22$.
5. Subtract: $36-22=14$.
6. Hence ${\alpha }^2+{\beta }^2+{\gamma }^2=14$.

Concept tested: Relations between roots and coefficients

Published 21 August 2026 · How this solution was prepared and checked

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