If in the expansion (1+x)m(1−x)n, the coefficient of x and x2 are 3 and -6 respectively then m is

We have, (1+x)m(1−x)n  = mC0+mC1x+mC2x2+…+mCmxm× nC0−nC1x+nC2x2…..+(−1)nnCnxn=mC0nC0− mC0nC1−nC0mC1x+ mC0nC2+nC0mC2−mC1⋅nC1x2+…..It is given that the coefficients of x and  x 2 in the expansion (1+x)m(1−x)m are 3 and -6 respectively  ∴− mC0nC1−nC0mC1=3and ,  mC0nC2+nC0mC2−mC1nC1=−6 ⇒m−n=3 and n(n−1)+m(m−1)−2mn=−12 ⇒m−n=3 and (m−n)2−(m+n)=−12 ⇒m−n=3 and  m+n=21 ⇒m=12,n=9