\(a)4{{\rm{x}}^2} - 12{\rm{x}}y + 9{y^2} = {\left( {2{\rm{x}}} \right)^2} - 2.2{\rm{x}}.3y + {\left( {3y} \right)^2} = {\left( {2{\rm{x}} - 3y} \right)^2}\)
\(b){x^3} + 9{{\rm{x}}^2} + 27{\rm{x}} + 27 = {x^3} + 3.{x^2}.3 + 3.x{.3^2} + {3^3} = {\left( {x + 3} \right)^3}\)
\(c)8{y^3} - 12{y^2} + 6y - 1 = {\left( {2y} \right)^3} - 3.{\left( {2y} \right)^2}.1 + 3.2y{.1^2} - {1^3} = {\left( {2y - 1} \right)^3}\)
\(\begin{array}{l}d) {\left( {2{\rm{x}} + y} \right)^2} - 4{y^2}\\ = {\left( {2{\rm{x}} + y} \right)^2} - {\left( {2y} \right)^2}\\ = \left( {2{\rm{x}} + y + 2y} \right)\left( {2{\rm{x}} + y - 2y} \right) = \left( {2{\rm{x}} + 3y} \right)\left( {2{\rm{x}} - y} \right)\end{array}\)
\(e) 27{y^3} + 8 = {\left( {3y} \right)^3} + {2^3} = \left( {3y + 2} \right)\left( {9{y^2} - 6y + 4} \right)\)
\(g) 64 - 125{{\rm{x}}^3} = {4^3} - {\left( {5{\rm{x}}} \right)^3} = \left( {4 - 5{\rm{x}}} \right)\left( {16 + 20{\rm{x}} + 25{{\rm{x}}^2}} \right)\)