Sum of cubes of the roots

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Let $x_{1}$ and $x_{2}$ be the roots of $x^{2} + p x + q = 0$ . Express $x_{1}^{3} + x_{2}^{3}$ in terms of $p$ and $q$ without solving the equation.

Posed by School algebra, after François Viète

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algebra polynomials

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x_{1}^{3} + x_{2}^{3} = 3 pq - p^{3}

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By Vieta's formulas $x_{1} + x_{2} = - p$ and $x_{1} x_{2} = q$ . Then $x_{1}^{3} + x_{2}^{3} = (x_{1} + x_{2})^{3} - 3 x_{1} x_{2} (x_{1} + x_{2}) = - p^{3} + 3 pq$ .

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Vieta's formulas

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