How do you simplify #(x+3)^2/(x+3)^3#?

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Answer 1 · 2016-05-07T10:42:55

We have that

#(x+3)^2/(x+3)^3=(x+3)^2/[(x+3)^2*(x+3)]= cancel((x+3)^2)/[cancel((x+3)^2)*(x+3)]=1/(x+3)#

Assuming that #x!=-3#

Answer 2 · 2016-05-07T10:50:23

#1/(x+3)#

#(x+3)^2/(x+3)^3#

#=(x+3)^2/(x+3)^(2+1)#

#=(x+3)^2/[(x+3)^2xx(x+3)]# because, #color(red)(a^(n+m)=a^nxxa^m#

#=cancel((x+3)^2)/[cancel((x+3)^2)xx(x+3)]#

#=1/(x+3)#

Answer 3 · 2016-05-07T11:01:52

#1/(x+3)#

Notice that the bases are the same. It does not matter how simple or how complicated the bases are, as long as they are the SAME.

You will do it in the same way as #x^2/x^3#.

Either: #(x*x)/(x*x*x)# = #1/x#

Or, by subtracting the indices:

#1/x^(3-2) = 1/x#

For the bases as #(x+3)#, the same applies:

#1/(x+3)^(3-2) = 1/(x+3)#