Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.8 AIME Find the answer

1. The roots $x_{1}, x_{2}$ of the equation $x^{2}-a x-a=0$ satisfy the relation $x_{1}{ }^{3}+x_{2}{ }^{3}+x_{1}{ }^{3} x_{2}{ }^{3}=75$ . Then $1993+5 a^{2}$ $+9 a^{4}=$ $\qquad$

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

1. According to Vieta's formulas, we have $x_{1}^{3}+x_{2}^{3}+x_{1}^{3} x_{2}^{3}=75$ . Thus, $a^{2}=25, a^{4}=625$ . Therefore, $1993+5 a^{2}+9 a^{4}=7743$ .

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