Đáp án:
\(\begin{array}{l}
a,\\
A = - 1\\
b,\\
B = 1\\
c,\\
C = 485
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
A = 2\left( {{x^3} + {y^3}} \right) - 3\left( {{x^2} + {y^2}} \right)\\
= \left[ {2\left( {{x^3} + {y^3}} \right) + 6xy} \right] + \left[ { - 3\left( {{x^2} + {y^2}} \right) - 6xy} \right]\\
= 2.\left( {{x^3} + {y^3} + 3xy} \right) - 3.\left( {{x^2} + {y^2} + 2xy} \right)\\
= 2.\left[ {{x^3} + {y^3} + 3xy\left( {x + y} \right)} \right] - 3.\left( {{x^2} + 2xy + {y^2}} \right)\,\,\,\,\,\,\,\,\left( {x + y = 1} \right)\\
= 2.\left( {{x^3} + 3{x^2}y + 3x{y^2} + {y^3}} \right) - 3.{\left( {x + y} \right)^2}\\
= 2.{\left( {x + y} \right)^3} - 3.{\left( {x + y} \right)^2}\\
= {2.1^3} - {3.1^2}\\
= - 1\\
b,\\
B = {x^3} + {y^3} + 3xy\\
= {x^3} + {y^3} + 3xy.1\\
= {x^3} + {y^3} + 3xy\left( {x + y} \right)\\
= {x^3} + {y^3} + 3{x^2}y + 3x{y^2}\\
= {x^3} + 3{x^2}y + 3x{y^2} + {y^3}\\
= {\left( {x + y} \right)^3}\\
= {1^3} = 1\\
c,\\
C = 8{x^3} - 27{y^3}\\
= {\left( {2x} \right)^3} - {\left( {3y} \right)^3}\\
= \left( {2x - 3y} \right).\left[ {{{\left( {2x} \right)}^2} + 2x.3y + {{\left( {3y} \right)}^2}} \right]\\
= \left( {2x - 3y} \right).\left\{ {\left[ {{{\left( {2x} \right)}^2} - 2.2x.3y + {{\left( {3y} \right)}^2}} \right] + 3.2x.3y} \right\}\\
= \left( {2x - 3y} \right).\left[ {{{\left( {2x - 3y} \right)}^2} + 18xy} \right]\\
= 5.\left( {{5^2} + 18.4} \right)\\
= 485
\end{array}\)
