Đáp án:
C=1320
Giải thích các bước giải:
\(\begin{array}{l}
Vi-et:\left\{ \begin{array}{l}
{x_1} + {x_2} = 10\\
{x_{1.}}{x_2} = - 8
\end{array} \right.\\
C = \left( {{x_1} - {x_2}} \right).\left( {{x_1} - {x_2}} \right)\left( {{x_1} + {x_2}} \right)\\
= {\left( {{x_1} - {x_2}} \right)^2}.\left( {{x_1} + {x_2}} \right)\\
= \left( {{x_1}^2 - 2{x_1}{x_2} + {x_2}^2} \right)\left( {{x_1} + {x_2}} \right)\\
= \left[ {\left( {{x_1}^2 + 2{x_1}{x_2} + {x_2}^2} \right) - 2{x_1}{x_2} - 2{x_1}{x_2}} \right]\left( {{x_1} + {x_2}} \right)\\
= \left[ {{{\left( {{x_1} + {x_2}} \right)}^2} - 4{x_1}{x_2}} \right]\left( {{x_1} + {x_2}} \right)\\
= \left( {{{10}^2} - 4.\left( { - 8} \right)} \right).10\\
= \left( {100 + 32} \right).10 = 132.10 = 1320
\end{array}\)
