Question #69c6c

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Answer 1 · 2017-04-08T16:41:27

I get #-200#

Expand #(x-2/x)^6 = (x-2x^-1)^6# to get

#x^6 - 6x^5(2x^-1)+15x^4(4x^-2)-20x^3(8x^-3)+15x^3(16x^-4) - 6x^2(32x^-5)+(64x^-6)#

When we distribute the #2# from #(2+3/x^2)# the product with every term except #-20x^3(8x^-3)# will depend on #x#. That product is

#2 xx -20 xx 8 = -320#

And when we distribute the #3/x^2# the product of #3/x^2# and #15x^4(4x^-2)# will be independent of #x#. That product is

#2 xx 15 xx 4 = 120#

The sum of those two products is

#-200#