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

3 Answers

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#

May 7, 2016

Answer:

#1/(x+3)#

Explanation:

#(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)#

May 7, 2016

Answer:

#1/(x+3)#

Explanation:

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)#