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Answer 1 · 2017-04-27T22:00:39

#(e^2 - 2e -1)/e#

First, let's discuss #e^1 and e^-1#

#e^1# just means #e#
#e^-1# means #1/e#
[If there was #e^-2# it would have meant #1/e^2#]

So now it becomes really simple.

Also #(-1)^3 = -1#
[#-1# raised to any odd number is #-1# and if it is raised to any even number, it is #1#]

Let's solve this

#(e-1/3) - (1/e +1/3)#
#e-1/3-1/e-1/3#

This becomes #(3e^2 - 2e -3)/(3e)#