Đáp án:
$\begin{array}{l}
36)8{x^3} - \dfrac{1}{{27}}{y^3}\\
= {\left( {2x} \right)^3} - {\left( {\dfrac{1}{3}y} \right)^3}\\
= \left( {2x - \dfrac{1}{3}y} \right)\left( {4{x^2} + 2x.\dfrac{1}{3}y + \dfrac{1}{9}{y^2}} \right)\\
= \left( {2x - \dfrac{1}{3}y} \right)\left( {4{x^2} + \dfrac{2}{3}xy + \dfrac{1}{9}{y^2}} \right)\\
37)\\
27{c^3} + \dfrac{8}{{125}}{z^3}\\
= {\left( {3c} \right)^3} + {\left( {\dfrac{2}{5}z} \right)^3}\\
= \left( {3c + \dfrac{2}{5}z} \right)\left( {9{c^2} - \dfrac{6}{5}cz + \dfrac{3}{{25}}{z^2}} \right)\\
38)\\
8{x^6} + 27{x^3}{y^9}\\
= {\left( {2{x^2}} \right)^3} + {\left( {3x{y^3}} \right)^3}\\
= \left( {2{x^2} + 3x{y^3}} \right)\left( {4{x^4} - 6{x^3}{y^3} + 9{x^2}{y^6}} \right)\\
39)\\
36{c^2} + 6cb + \dfrac{1}{4}{b^2} = {\left( {6c + \dfrac{1}{2}b} \right)^2}\\
40)\\
\dfrac{4}{{25}}{t^4}{m^2} - \dfrac{1}{{169}}{c^6}{z^{10}}\\
= \left( {\dfrac{2}{5}{t^2}m - \dfrac{1}{{13}}{c^3}{z^5}} \right)\left( {\dfrac{2}{5}{t^2}m + \dfrac{1}{{13}}{c^3}{z^5}} \right)
\end{array}$
