The coefficient of x^(-5)=?, in the binomial expansion of [((x+1)/((x^(2/3)-(x^(1/3))+1))-((x-1)/(x-(x^(1/2))]^10 , where 'x' not equal to 0,1 is:

enter image source here

1 Answer
Feb 23, 2018

1

Explanation:

Since #a^3+1 = (a+1)(s^2-a+1)#, we have
#x+1 = (x^{1/3}+1)(x^{2/3}-x^{1/3}+1)#, so that
#{x+1}/{x^{2/3}-x^{1/3}+1}=x^{1/3}+1#
Again
#{x-1}/{x-x^{1/2}}={(x^{1/2}-1)(x^{1/2}+1)}/{x^{1/2}(x^{1/2}-1)}= 1+x^{-1/2}#

Thus the given expression simplifies to

#(x^{1/3}-x^{-1/2})^10#

The only way that we can get #x^{-5}# here is from the term
#.^10C_10(x^{1/3})^0 (-x^{-1/2})^10# - and the coefficient of this term is #(-1)^10 = 1#