Phân tích các đa thức sau thành nhân tử:
a)\({x^3} + \frac{1}{{27}};\)
b) \({(a + b)^3} - {(a - b)^3};\)
c) \({(a + b)^3} + {(a - b)^3};\)
d) \(8{x^3} + 12{x^2}y + 6x{y^2} + {y^3};\)
e) \( - {x^3} + 9{x^2} - 27x + 27;\)
Câu a:
\(\begin{array}{l} {x^3}\; + \;\frac{1}{{27}}\;\\ = {x^3}\; + {(\frac{1}{3})^3}\;\\ = (x + \;\frac{1}{3})({x^2}-x.\frac{1}{3} + {(\frac{1}{3})^2})\\ = (x + \;\frac{1}{3})({x^2}-\frac{1}{3}x + \;\frac{1}{9}) \end{array}\)
Câu b:
\(\begin{array}{*{20}{l}} {{{\left( {a + b} \right)}^3}-{{\left( {a - b} \right)}^3}\;\;{\rm{ }}\;}\\ { = \left[ {\left( {a + b} \right)-\left( {a-b} \right)} \right]\left[ {{{\left( {a + b} \right)}^2} + \left( {a + b} \right).\left( {a-b} \right) + {{\left( {a-b} \right)}^2}} \right]}\\ { = \left( {a + b-a + b} \right)\left( {{a^2} + 2ab + {b^2} + {a^2}-{b^2} + {a^2}-2ab + {b^2}} \right)}\\ { = {\rm{ }}2b\left( {3{a^3} + {b^2}} \right)} \end{array}\)
Câu c:
\(\begin{array}{*{20}{l}} \begin{array}{l} {\left( {a + b} \right)^3} + {\left( {a-b} \right)^3}\;\\ = \left[ {\left( {a + b} \right) + \left( {a-b} \right)} \right]\left[ {{{\left( {a + b} \right)}^2}-\left( {a + b} \right)\left( {a-b} \right) + {{\left( {a-b} \right)}^2}} \right] \end{array}\\ { = \left( {a + b + a-b} \right)\left( {{a^2} + 2ab + {b^2}-{a^2}\; + {b^{2}} + {a^2}-2ab + {b^2}} \right]}\\ { = 2a.\left( {{a^2} + 3{b^2}} \right)} \end{array}\)
Câu d:
\(\begin{array}{l} 8{x^3}\; + {\rm{ }}12{x^2}y + 6x{y^2}\; + {\rm{ }}{y^3}\;\\ = {\left( {2x} \right)^3} + 3.{\left( {2x} \right)^2}.y\; + 3.2x.y + {y^3}\;\\ = {\left( {2x + y} \right)^3} \end{array}\)
Câu e:
\(\begin{array}{l} - {x^{3}} + 9{x^2}-27x + 27\\ = 27-27x + 9{x^2}-{x^3}\\ = {3^3}-{3.3^2}.x + 3.3.{x^2}-{x^3}\\ = {\left( {3-x} \right)^3} \end{array}\)
-- Mod Toán 8
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