Question

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Answer 1

Ответ:

$335$

Объяснение:

$x^{2}-5x-14=0;$

$\displaystyle \left \{ {{x_{1}+x_{2}=-(-5)} \atop {x_{1} \cdot x_{2}=-14}} \right. \Leftrightarrow \left \{ {{x_{1}+x_{2}=5} \atop {x_{1} \cdot x_{2}=-14}} \right. ;$

$x_{1}^{3}+x_{2}^{3}=(x_{1}+x_{2})(x_{1}^{2}-x_{1} \cdot x_{2}+x_{2}^{2});$

$x_{1}^{2}+x_{2}^{2}=x_{1}^{2}+2 \cdot x_{1} \cdot x_{2}+x_{2}^{2}-2 \cdot x_{1} \cdot x_{2}=(x_{1}+x_{2})^{2}-2 \cdot x_{1} \cdot x_{2};$

$x_{1}^{3}+x_{2}^{3}=(x_{1}+x_{2})((x_{1}+x_{2})^{2}-2 \cdot x_{1} \cdot x_{2}-x_{1} \cdot x_{2});$

$x_{1}^{3}+x_{2}^{3}=(x_{1}+x_{2})((x_{1}+x_{2})^{2}-3 \cdot x_{1} \cdot x_{2});$

$x_{1}^{3}+x_{2}^{3}=5 \cdot (5^{2}-3 \cdot (-14))=5 \cdot (25+42)=125+210=335;$