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(x^3 -3x)^6. find the term in x^12
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In the expansion of (x+a)^n \[general~ term ~T_{r+1}=C_{r}^n{x}^{n-r}a^r\] calculate general term.
write x^3 in place of x and -3x in place of a. n=6
\[T _{r+1}=C_{r}^{6}\left( x^3 \right)^{6-r}\left( -3x \right)^r =C _{r}^{6}x ^{18-3r}\left( -1 \right)^r3^rx^r\] \[=\left( -1 \right)^r3^rC _{r}^{6}x ^{18-3r+r}=\left( -1 \right)^r3^rC _{r}^{6}x ^{18-2r}\] Put 18-2r=12 2r=18-12=6 r=3 \[C _{3}^{6}=\frac{ 6\times5\times4 }{ 3\times2\times1 }=20\] simplify and find the required term.
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