How do you simplify #(-91a^4b+14ab )/ (-7ab)#?

1 Answer
Apr 19, 2016

Answer:

#(-91a^4b+14ab)/(-7ab)#

#= 13a^3-2#

#=(root(3)(13)a-root(3)(2))(root(3)(169)a^2+root(3)(26)a+root(3)(4))#

with exclusions #a!=0# and #b!=0#

Explanation:

Both of the terms #-91a^4b# and #14ab# are divisible by #-7ab#, so we find:

#(-91a^4b+14ab)/(-7ab) = 13a^3-2#

with exclusions #a!=0# and #b!=0#

Note that the exclusions are required because if #a=0# or #b=0# then the original expression is of the form #0/0# which is undefined.

I guess that counts as 'simplified', but it can be factored too, using cube roots...

For any Real numbers #x# and #y# note that:

#root(3)(x) root(3)(y) = root(3)(xy)#

We will use this below.

The difference of cubes identity can be written:

#A^3-B^3 = (A-B)(A^2+AB+B^2)#

We can factor #13a^3-2# using this identity with #A=root(3)(13)a# and #b = root(3)(2)# as follows:

#13a^3-2#

#=(root(3)(13)a)^3-(root(3)(2))^3#

#=(root(3)(13)a-root(3)(2))((root(3)(13)a)^2+(root(3)(13)a)(root(3)(2))+(root(3)(2))^2)#

#=(root(3)(13)a-root(3)(2))(root(3)(169)a^2+root(3)(26)a+root(3)(4))#