\[\begin{array}{l}
{\left( {1 + 2x - 3{x^2}} \right)^8} = {\left[ {\left( {1 - x} \right)\left( {1 + 3x} \right)} \right]^8} = {\left( {1 - x} \right)^8}.{\left( {1 + 3x} \right)^8}\\
= \sum\limits_{k = 0}^8 {C_8^k{{\left( { - x} \right)}^k}} .\sum\limits_{l = 0}^8 {C_8^l{{\left( {3x} \right)}^l}} = \sum\limits_{k = 0}^8 {C_8^k{{\left( { - 1} \right)}^k}{x^k}} .\sum\limits_{l = 0}^8 {C_8^l{{.3}^l}{x^l}} \\
= \sum\limits_{k = 0}^8 {\sum\limits_{l = 0}^8 {C_8^k{{\left( { - 1} \right)}^k}C_8^l{3^l}} {x^{k + l}}} \\
Cho\,k + l = 9 = 1 + 8 = 2 + 7 = 3 + 6 = 4 + 5\\
= 5 + 4 = 6 + 3 = 7 + 2 = 8 + 1\\
TH1:k = 1,l = 8 \Rightarrow he\,so\,C_8^1.{\left( { - 1} \right)^1}.C_8^8{.3^8} = - {8.3^8}\\
TH2:k = 2,l = 7 \Rightarrow he\,so\,C_8^2.{\left( { - 1} \right)^2}.C_8^7{.3^7} = C_8^2.C_8^7{.3^7}\\
...
\end{array}\]
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