How do you evaluate #a(2(a^2-a)-a(a(a+1)/(a^2+a))-(-a)^2-(a^2-a)/(a(a-1)))#?

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Answer 1 · 2017-10-22T16:09:57

# = a(a^2 -3a - 1) = a^3 -3a^2 -a#

We will simplify term by term and then combine them.

#(2(a^2-a) = 2a ^2 - 2a color (white)(aaa) (1)#

#a(a ((a+1) / (a^2+a) ) = a ( a( cancel (a + 1) / (a cancel(a+ 1))))#
# = ( a * cancel a)/ cancel a = a color (white)(aaa) (2)#

# (-a)^2 = a^2 color (white)(aaa) (3)#

#(a^2 - a) / (a (a - 1)) = cancel (a (a - 1)) / cancel (a (a -1) )= 1 color (white)(aaa) #(4)

Combining all the terms,

# a ( 2a ^2 - 2a - a - a^2 - 1)#

# = a ( a^2 - 3a - 1) #

# = a^3 - 3a^2 - a#