Đáp án:
$1)\left( {x + 1} \right)\left( {{x^2} + x + 1} \right)\left( {x - 1} \right)\left( {{x^2} - x + 1} \right) = {x^6} - 1$
$2){\left( {x - 2} \right)^3} - \left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) + 6{\left( {x + 1} \right)^2} = 9{x^2} - 3x + 25$
$3){\left( {x + 3} \right)^3} - {\left( {x + 1} \right)^3} = 2\left( {3{x^2} + 12x + 13} \right)$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
B{C^2} = A{B^2} + A{C^2} - AB.AC\\
AB = c;AC = b;BC = a\\
\Leftrightarrow {a^2} = {b^2} + {c^2} - bc\\
1)\left( {x + 1} \right)\left( {{x^2} + x + 1} \right)\left( {x - 1} \right)\left( {{x^2} - x + 1} \right)\\
= \left[ {\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)} \right]\left[ {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)} \right]\\
= \left( {{x^3} + 1} \right)\left( {{x^3} - 1} \right)\\
= {x^6} - 1\\
2){\left( {x - 2} \right)^3} - \left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) + 6{\left( {x + 1} \right)^2}\\
= {\left( {x - 2} \right)^3} - {\left( {x - 3} \right)^3} + 6{\left( {x + 1} \right)^2}\\
= \left( {\left( {x - 2} \right) - \left( {x - 3} \right)} \right)\left[ {{{\left( {x - 2} \right)}^2} + \left( {x - 2} \right)\left( {x - 3} \right) + {{\left( {x - 3} \right)}^2}} \right] + 6{\left( {x + 1} \right)^2}\\
= {\left( {x - 2} \right)^2} + \left( {x - 2} \right)\left( {x - 3} \right) + {\left( {x - 3} \right)^2} + 6{\left( {x + 1} \right)^2}\\
= {x^2} - 4x + 4 + {x^2} - 5x + 6 + {x^2} - 6x + 9 + 6{x^2} + 12x + 6\\
= 9{x^2} - 3x + 25\\
3){\left( {x + 3} \right)^3} - {\left( {x + 1} \right)^3}\\
= \left( {\left( {x + 3} \right) - \left( {x + 1} \right)} \right)\left[ {{{\left( {x + 3} \right)}^2} + \left( {x + 3} \right)\left( {x + 1} \right) + {{\left( {x + 1} \right)}^2}} \right]\\
= 2\left( {{{\left( {x + 3} \right)}^2} + \left( {x + 3} \right)\left( {x + 1} \right) + {{\left( {x + 1} \right)}^2}} \right)\\
= 2\left( {{x^2} + 6x + 9 + {x^2} + 4x + 3 + {x^2} + 2x + 1} \right)\\
= 2\left( {3{x^2} + 12x + 13} \right)
\end{array}$
