How do you factor this, it is complicated long task and i know the answer but i am interested if there is shorter way to solution?

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Answer 1 · 2018-01-22T06:28:03

#(2/(a^b-4b)-b/(a^3-4a)-1/(a^2+2a)):(b^2-4b+4)/(a^3b-4ab)=(a-b^2+2):(b-2)^2#

First factorize all binomials and trinomials i.e.

#(2/(a^b-4b)-b/(a^3-4a)-1/(a^2+2a)):(b^2-4b+4)/(a^3b-4ab)#

or #(2/(b(a-2)(a+2))-b/(a(a-2)(a+2))-1/(a(a+2))):(b-2)^2/(ab(a-2)(a+2)#

multiplying both by #ab(a-2)(a+2)#, we get

#(2a-b^2-(a-2)):(b-2)^2#

or #(a-b^2+2):(b-2)^2#

Answer 2 · 2018-01-22T06:48:31

# (a+b):(2-b)#.

We have, the Expression

#={2/(b(a+2)(a-2))-b/(a(a+2)(a-2))-1/(a(a+2))}:(b-2)^2/(ab(a^2-4))#,

#={2a-b^2-b(a-2)}/cancel{ab(a+2)(a-2)}:(b-2)^2/cancel(ab(a+2)(a-2))#,

#={2a-b^2-ab+2b}:(b-2)^2#,

#={ul(2a+2b)-ul(ab-b^2)}:(b-2)^2#,

#={2(a+b)-b(a+b)}:(b-2)^2#,

#={(a+b)(2-b)}:(2-b)^2......[because, (b-2)^2=(2-b)^2]#,

#rArr" The Exp.="(a+b):(2-b)#.