Đáp án:
$\begin{array}{l}
g)\dfrac{1}{{27}}{x^3} + {x^2} + 9x + 27\\
= {\left( {\dfrac{1}{3}x} \right)^3} + 3.\dfrac{1}{9}{x^2}.3 + 3.\dfrac{1}{3}x{.3^2} + {3^3}\\
= {\left( {\dfrac{1}{3}x + 3} \right)^3}\\
h)8{u^3} - 60{u^2}v + 150u{v^2} - 125{v^3}\\
= {\left( {2u} \right)^3} - 3.4{u^2}.5v + 3.2u.25{v^2} - {\left( {5v} \right)^3}\\
= {\left( {2u - 5v} \right)^3}\\
i){x^3} + 3{x^2} + 3x + 1\\
+ 3\left( {{x^2} + 2x + 1} \right)y + 3x{y^2} + 3{y^2} + {y^3}\\
= {\left( {x + 1} \right)^3} + 3.{\left( {x + 1} \right)^2}.y + 3.\left( {x + 1} \right).{y^2} + {y^3}\\
= {\left( {x + 1 + y} \right)^3}
\end{array}$
