How do you solve 3 over b^3 + 3 over b^2 equals 6 over b?

Question

anonymous · Answer

please use the equation editor so that i may understand your question clearly. :)

anonymous · Answer

3/(b^3+3)/b^2=6/b?
Then it becomes 3b^2/(b^3+3)=6/b
==> 3b^3/(b^3+3)=6
==> 3b^3=6(b^3+3)
==> 3b^3=6b^3+18
==> 6b^3-3b^3=-18
==> 3b^3=-18
==> :. b^3=-6
Hope this helps...

anonymous · Answer

3          3       6
---  + --- = --
b^3    b^2      b

anonymous · Answer

get the lcd, in this case is b^3 and multiple b^3 to the whole equation to get

3+3b=6b^2
6b^2-3b-3=0
2b^2-b-1
(2b+1)(b-1)=0
b=-1/2, b=1

hopefully i didn't have any mistakes there!

anonymous · Answer

How is it negative?

anonymous · Answer

(2b+1)(b-1)=0 
equate each to 0
2b+1=0
2b=-1
b==1/2

b-1=0
b=1 

:)

anonymous · Answer

i meant, b=-1/2 there. whoops!