Ta có \({\left( {x + a} \right)^3}{\left( {x - b} \right)^6} \)
\(= \sum\limits_{m = 0}^3 {C_3^m{x^{3 - m}}{a^m}} \sum\limits_{n = 0}^6 {C_6^n{x^{6 - n}}{{( - b)}^n}}\)
\(= \left( {C_3^0{x^3} + C_3^1{x^2}a + C_3^2x{a^2} + C_3^3{a^3}} \right)\)
\([ C_6^0{x^6} + C_6^1{x^5}( - b) + C_6^2{x^4}{{(-b)}^2} + \)
\(C_6^3{x^3}{{( - b)}^3} + C_6^4{x^2}{{(-b)}^4} + C_6^5x{{( - b)}^5} \)
\(+ C_6^6{{(-b)}^6}] \)
Số hạng chứa \({x^7}\) là \([C_3^0.C_6^2{(- b)}^2 +C_3^1a.C_6^1{( - b)}\)
\(+ C_3^2a^2C_6^0 ]x^7\)
Số hạng chứa \({x^8}\) là \(\left[ {C_3^0.C_6^1\left( { - b} \right) + C_3^1a.C_6^0} \right]{x^8}\).
Theo bài ra ta có
\(\left\{ \begin{array}{l}15{b^2} - 18ab + 3{a^2} = - 9\\ - 6b + 3a = 0\end{array} \right. \)
\(\Rightarrow \left\{ \begin{array}{l}a = 2b\\{b^2} = 1\end{array} \right.\)
\(\Rightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}a = 2\\b = 1\end{array} \right.\\\left\{ \begin{array}{l}a = - 2\\b = - 1.\end{array} \right.\end{array} \right.\)
