Let, y=(1−x)2008(1+x+x2)2007
⇒y=(1−x)(1−x)2007(1+x+x2)2007
⇒y=(1−x)[(1−x)(1+x+x2)]2007
⇒y=(1−x)(1−x3)2007
⇒y=(1−x)(C02007−C12007(x3)+......)
General term in the expansion is given by,
Tr+1=(1−x)(Cr2007x3r(−1)r)
⇒Tr+1=2007Crx3r(−1)r−2007Crx3r+1(−1)r
For coefficient of x12,
3r=2012
But, r=32012
Also, 3r+1=2012
⇒3r=2011
But, r=32011
Hence, there is no term containing x2012.
So coefficient of x2012=0