# Simplify `(4(x^2 - 9)^2 - (x+3)^2) / (x^2 + 6x + 9)`

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You may should notice the difference of squares to numerator, hence you may use a special product instead of the difference of squares such that:

`4(x^2 - 9)^2 - (x+3)^2 = (2(x^2 - 9) - (x+3))(2(x^2 - 9) + (x + 3))`

Notice the difference of squares `x^2 - 9` that may be substituted by `(x-3)(x+3)` such that:

`(2(x-3)(x+3) - (x+3))(2(x-3)(x+3)+ (x+3))`

Factoring out x+3 yields:

`(x+3)^2*(2(x-3)-1)(2(x-3)+1)`

Write the new form of the fraction such that:

`((x+3)^2*(2(x-3)-1)(2(x-3)+1))/(x^2 + 6x + 9)`

You should notice that the denominator is expansion of a squared binomial such that:

`x^2 + 6x + 9 = (x + 3)^2`

`` `((x+3)^2*(2(x-3)-1)(2(x-3)+1))/((x + 3)^2)`

Reducing by `(x + 3)^2` yields:

`(2(x-3)-1)(2(x-3)+1)) = 4(x-3)^2 - 1`

**Hence, simplifying the fraction yields `4(x-3)^2 - 1` .**