Perform indicated operation?

#(2p)/(4p^2-1)/(6p^3)/(6p+3)#

1 Answer
Jun 20, 2018

#((2p)/(4p^2-1))/((6p^3)/(6p+3))# = #1/(p^2(2p-1))#

Explanation:

#((2p)/(4p^2-1))/((6p^3)/(6p+3))#

This would be the same as multiplying the numerator
#(2p)/(4p^2-1)#
with the inverse of the denominator
#(6p^3)/(6p+3)#, i.e. #(6p+3)/(6p^3)#

We, therefore, get:
#((2p)/(4p^2-1))/((6p^3)/(6p+3))#
= #((2p)/(4p^2-1))*((6p+3)/(6p^3))# = #(2p(6p+3))/(6p^3(4p^2-1))#
= #(3(2p+1))/(3p^2(4p^2-1))# = #((2p+1))/(p^2(4p^2-1))#

Now we need to remember that (a+b)(a-b)=a^2-b^2
Therefore (2x+1)(2x-1)=(4x^2-1)

We can, therefore, write:
#((2p+1))/(p^2(2p+1)(2p-1))=1/(p^2(2p-1))#