Question

((x^4-b^4)/(x^2-2bx+b^2))÷((x^2+bx)/(x-b))

Answer 1

the first thing you should do is change the divide to multiply by the inverse.
that means "flip" the 2nd fraction and multiply

Answer 2

so re-write the problem, multiplying by the inverse of the 2nd fraction

then factor the terms and look for factors that will cancel (divide into each other)
can you do that?

Answer 3

try dividing x-b into it

Answer 4

@phi i have (x^5-(b^4)x-bx^4+b^5)/((b^3)x-b^2x^2-bx^2+x^3)

Answer 5

Having trouble finding the factors

Answer 6

Here is what you started with
((x^4-b^4)/(x^2-2bx+b^2))÷((x^2+bx)/(x-b))
\[ \frac{x^4-b^4}{x^2-2bx+b^2}÷\frac{x^2+bx}{x-b}\]
Very first step is to flip the 2nd fraction and multiply:
\[ \frac{x^4-b^4}{x^2-2bx+b^2}\cdot \frac{x-b}{x^2+bx}\]

the x^4-b^4 is a difference of squares: (x^2)^2 - (b^2)^2
the term x^2+bx in the bottom can have an x factored out x(x+b)

x-b does go evenly into x^2-2bx+b^2
I would use that info to rewrite this mess, and see what is what before going on...

Answer 7

@phi how i have (x^2-b^2)(x^2+b^2)/(x-b)(x-b)÷(x-b)/x(x+b)

Answer 8

And i believe that x^2-b^2 can be factored further

Answer 9

I hope you mean
(x^2-b^2)(x^2+b^2)(x-b)/(x-b)(x-b)x(x+b)
or
\[ \frac{(x^2-b^2)(x^2+b^2)(x-b)}{(x-b)(x-b)x(x+b)}\]

yes you can factor x^2-b^2
but right away we can cancel (x-b) from top and bottom

Answer 10

Yes

Answer 11

Will x^2-b^2 = (x-b)(x+b)

Answer 12

Which will cancel with x(x+b) leaving 1/x

Answer 13

On the rhs

Answer 14

yes, it is a difference of squares. if you use FOIL you can check:
(x-b)(x+b)= x^2 +bx -bx -b^2 = x^2-b^2

Answer 15

before any cancelling this is what you should get
\[ \frac{(x-b)(x+b)(x^2+b^2)(x-b)}{(x-b)(x-b)x(x+b)}\]

Answer 16

Yes

Answer 17

I assume you can get the answer from here. The only thing to add is that you should exclude x= b , x=-b and x=0 as these would cause divide by 0 in the original expression.

Answer 18

Yes thank you phi