if `(1-x^3)^n=sum_(r=0)^n a_rx^r (1-x)^(3n-2r), where n epsilonN ` then find `a_r`.

Explanation

We have , ` (1 - x^(3))^(n) = sum_(r=0)^(n) a_(r) x^(r) (1 - x)^(3n-2r)`
` rArr (1 - x)^(n) (1 + x + x^(2))^(n) = sum_(r=0)^(n) (a_(r) *x^(r) (1-x)^(3n))/((1 - x)^(2r))`
` rArr ((1 - x)^(n) (1 +x+x^(2))^(n))/((1 + x)^(3n)) = sum_(r=0)^(n) (a_(r) *x^(r))/((1 - x)^(2r))`
`rArr [(1 + x+ x^(2))/((1 -x)^(2))]^(n) = sum_(r=0)^(n)a_(r) * (x^(r))/((1 - x)^(2r))`
` rArr [((1 - x)^(2) + 3x)/((1 - x)^(2))]^(n) = sum_(r=0)^(n) a_(r) [(x)/((1 - x)^(2))]`
` rArr [1+ 3((x)/((1 - x)^(2)))]^(n) = sum_(r=0)^(n) a_(r) [(x)/((1 - x)^(2))]`...(i)
Let ` A = (x)/((1 - x)^(2))`
Then , Eq .(i) becomes ` (1 + 3A)^(n) = sum_(r=0)^(n) a_(r) A^(r)`
on comparing the coefficient of ` A^(r)` , we get
` ""^(n)C_(r) * 3^(r) = a_(r)`
Hence ` a_(r) = ""^(n)C_(r) * 3^(r)` .