For binomial expansions of the form (x+y)^n(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)
