Phân tích đa thức thành nhân tử: a,x^4+2x^3-6x-9 b,(a^2+b^2+ab)^2-a^2b^2-b^2c^2-c^2a^2 c,x^2(y^2-4)-6y(y^2-4)+9 d,a^4-9a^3+81a-81

Giải thích các bước giải:

$\begin{array}{l}
a){x^4} + 2{x^3} – 6x – 9\\
 = \left( {{x^4} – 9} \right) + \left( {2{x^3} – 6x} \right)\\
 = \left( {{x^2} – 3} \right)\left( {{x^2} + 3} \right) + 2x\left( {{x^2} – 3} \right)\\
 = \left( {{x^2} – 3} \right)\left( {{x^2} + 3 + 2x} \right)\\
 = \left( {{x^2} – 3} \right)\left( {{x^2} + 2x + 3} \right)\\
b){\left( {{a^2} + {b^2} + ab} \right)^2} – {a^2}{b^2} – {b^2}{c^2} – {c^2}{a^2}\\
 = {\left( {{a^2} + {b^2} + ab} \right)^2} – {\left( {ab} \right)^2} – {c^2}\left( {{b^2} + {a^2}} \right)\\
 = \left( {{a^2} + {b^2} + ab – ab} \right)\left( {{a^2} + {b^2} + ab + ab} \right) – {c^2}\left( {{b^2} + {a^2}} \right)\\
 = \left( {{a^2} + {b^2}} \right)\left( {{a^2} + 2ab + {b^2}} \right) – {c^2}\left( {{a^2} + {b^2}} \right)\\
 = \left( {{a^2} + {b^2}} \right){\left( {a + b} \right)^2} – {c^2}\left( {{b^2} + {a^2}} \right)\\
 = \left( {{a^2} + {b^2}} \right)\left( {{{\left( {a + b} \right)}^2} – {c^2}} \right)\\
 = \left( {{a^2} + {b^2}} \right)\left( {a + b – c} \right)\left( {a + b + c} \right)\\
c){x^2}\left( {{y^2} – 4} \right) – 6y\left( {{y^2} – 4} \right) + 9\\
 = \left( {{y^2} – 4} \right)\left( {{x^2} – 6y} \right) + 9\\
d){a^4} – 9{a^3} + 81a – 81\\
 = \left( {{a^4} – 81} \right) – 9a\left( {{a^2} – 9} \right)\\
 = \left( {{a^2} – 9} \right)\left( {{a^2} + 9} \right) – 9a\left( {{a^2} – 9} \right)\\
 = \left( {{a^2} – 9} \right)\left( {{a^2} + 9 – 9a} \right)\\
 = \left( {{a^2} – 9} \right)\left( {{a^2} – 9a + 9} \right)
\end{array}$