For binomial expansions of the form #(x+y)^n#, we have:
#sum_(r=0)^(n)((n),(r))x^(n-r)y^r#
#((n),(r))=color(white)(0)^nC_(r)=(n!)/(r!(n-r)!)#
Notice in this form we have the powers of #x# descending.
We would then expect the power of #x# in the third term to be:
#x^6#
So:
#x^(8-r)=x^6=>r=2#
Using:
#((n),(r))#
#((8),(2))c^(8-2)(-d)^2#
#((8),(2))=(8!)/(2!(8-2)!)=(8xx7xx6xx5xx4xx3xx2xx1)/(2xx1xx6xx5xx4xx3xx2xx1)#
#=(8xx7xxcancel(6)xxcancel(5)xxcancel(4)xxcancel(3)xxcancel(2)xxcancel(1))/(2xx1xxcancel(6)xxcancel(5)xxcancel(4)xxcancel(3)xxcancel(2)xxcancel(1))=(8xx7)/(2xx1)=56/2=28#
#:.#
#28c^(8-2)(-d)^2=color(blue)(28c^6d^2)#
