Bài 44 trang 20 SGK Toán 8 tập 1

Phân tích các đa thức sau thành nhân tử:a) \({x^3} + \dfrac{1}{27}\);

b) \({\left( {a + b} \right)^3} - {\left( {a - b} \right)^3}\);

c) \({\left( {a + b} \right)^3} + {\left( {a - b} \right)^3}\);

d) \(8{x^3} + 12{x^2}y + 6x{y^2} + {y^3}\)

e) \( - {x^3} + 9{x^2} - 27x + 27.\)

\(\begin{array}{l} a)\,\,{x^3} + \dfrac{1}{{27}} = {x^3} + {\left( {\dfrac{1}{3}} \right)^3}\\ = \left( {x + \dfrac{1}{3}} \right)\left[ {{x^2} - \dfrac{1}{3}x + {{\left( {\dfrac{1}{3}} \right)}^2}} \right]\\ = \left( {x + \dfrac{1}{3}} \right)\left( {{x^2} - \dfrac{1}{3}x + \dfrac{1}{9}} \right) \end{array}\)

\(d)\,\,8{x^3} + 12{{\rm{x}}^2}y + 6{\rm{x}}{y^2} + {y^3}\\ = {\left( {2{\rm{x}}} \right)^3} + 3.{\left( {2{\rm{x}}} \right)^2}.y + 3.2{\rm{x}}.{y^2} + {y^3}\\ = {\left( {2{\rm{x}} + y} \right)^3}\)

\(\begin{array}{l}e)\; - {x^3} + 9{x^2} - 27x + 27 \\= 27 - 27x + 9{x^2} - {x^3}\\ = {3^3} - {3.3^2}.x + 3.3.{x^2} - {x^3}\\ = {\left( {3 - x} \right)^3}.\end{array}\)