Đáp án:
$\begin{array}{l}
{\left( {x + 1} \right)^3} - {\left( {x - 1} \right)^3} - 6{\left( {x - 1} \right)^2} = - 10\\
\Leftrightarrow {x^3} + 3{x^2} + 3x + 1 - {x^3} + 3{x^2} - 3x + 1\\
- 6\left( {{x^2} - 2x + 1} \right) = - 10\\
\Leftrightarrow 6{x^2} + 2 - 6{x^2} + 12x - 6 = - 10\\
\Leftrightarrow 12x = - 6\\
\Leftrightarrow x = - \dfrac{1}{2}\\
Vậy\,x = - \dfrac{1}{2}\\
a){\left( {x + y} \right)^3} - {\left( {x - y} \right)^3} - 2{y^3}\\
= {x^3} + 3{x^2}y + 3x{y^2} + {y^3}\\
- \left( {{x^3} - 3{x^2}y + 3x{y^2} - {y^3}} \right) - 2{y^3}\\
= 6{x^2}y + 2{y^3} - 2{y^3}\\
= 6{x^2}y\\
b){\left( {x + y + z} \right)^2} - 2\left( {x + y + z} \right)\left( {x + y} \right) + {\left( {x + y} \right)^2}\\
= {\left( {x + y + z - x - y} \right)^2}\\
= {z^2}\\
c)\left( {x - 2} \right)\left( {{x^2} - 2x + 4} \right)\left( {x + 2} \right)\left( {{x^2} + 2x + 4} \right)\\
= \left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)\\
= \left( {{x^3} - {2^3}} \right)\left( {{x^3} + {2^3}} \right)\\
= \left( {{x^3} - 8} \right)\left( {{x^3} + 8} \right)\\
= {\left( {{x^3}} \right)^2} - {8^2}\\
= {x^6} - 64
\end{array}$
