Explanation:
Hey, there!!
Given that,
![\frac{ {x}^(2) - 9 }{x + 3}](https://img.qammunity.org/2021/formulas/mathematics/college/mdxxwopwpjk4hebuz8933s7xr9grq0r1gq.png)
{ we can write (a^2-4) as (a^2 - 2^2) also as (x^2- 9) can be written as (x^2 - 3^2)}.
![\frac{ {x}^(2) - {3}^(2) }{x + 3}](https://img.qammunity.org/2021/formulas/mathematics/college/5w1c4u7kqn8cf3dr2tl9e8402pfjdq5f9i.png)
We have a^2-b^2= (a+b) (a-b), so keep same formula on it.
![((x + 3)(x - 3))/((x + 3))](https://img.qammunity.org/2021/formulas/mathematics/college/n928ww84lthib9il07i50uyfoz5b2u1xum.png)
(x+3) in numerator and denominator gets cancelled,
![(x - 3)](https://img.qammunity.org/2021/formulas/mathematics/college/e0o0tj8aarzeyoxpjzyc21h0na8qypinf1.png)
Therefore, (x-3) is the final value.
Hope it helps...