Đáp án:
$\begin{array}{l}
{\left( {x + 1} \right)^3} - x\left( {{x^2} + 3x + 5} \right) = 0\\
\Leftrightarrow {x^3} + 3{x^2} + 3x + 1 - {x^3} - 3{x^2} - 5x = 0\\
\Leftrightarrow - 2x + 1 = 0\\
\Leftrightarrow x = \dfrac{1}{2}\\
Vậy\,x = \dfrac{1}{2}\\
{\left( {x + 2} \right)^3} - \left( {{x^2} + 3} \right)\left( {x + 6} \right) = 16\\
\Leftrightarrow {x^3} + 6{x^2} + 12x + 8 - {x^3} - 6{x^2} - 3x - 18 = 16\\
\Leftrightarrow 9x = 26\\
\Leftrightarrow x = \dfrac{{26}}{9}\\
Vậy\,x = \dfrac{{26}}{9}\\
B2)\\
8{x^6} + 12{x^4}y + 6{x^2}{y^2} + {y^3}\\
= {\left( {2{x^2}} \right)^3} + 3.{\left( {2{x^2}} \right)^2}.y + 3.2{x^2}.{y^2} + {y^3}\\
= {\left( {2{x^2} + y} \right)^3}\\
{x^6} - 3{x^4} + 3{x^2} - 1 = {\left( {{x^2} - 1} \right)^3}\\
{\left( {x + y} \right)^3}{\left( {x - y} \right)^3} = {\left( {{x^2} - {y^2}} \right)^3}
\end{array}$
