Verify if the difference (1-9/y^2)/(1-3/y)-3/y=1

We'll multiply the second fraction by the term
(1-3/y):

[(1-9/y^2) -
(3/y)*(1-3/y)]/(1-3/y)

The difference of squares returns
the product:

1-9/y^2 =
(1-3/y)(1+3/y)

[(1-9/y^2) - (3/y)*(1-3/y)]/(1-3/y) =
[(1-3/y)(1+3/y) -
(3/y)

(1-3/y)]/(1-3/y)

We'll
factorize the numerator by (1-3/y):

(1-3/y) [(1+3/y) -
(3/y)]/(1-3/y)

We'll simplify by
(1-3/y):

(1-3/y) [(1+3/y) - (3/y)]/(1-3/y) =  [(1+3/y) -
(3/y)]

We'll eliminate like terms inside
brackets:

(1-9/y^2)/(1-3/y)-3/y=1

We
notice that the given difference yields 1.