How do 3 fractions having respectively $(a-x)^2$ , $(a+x)^2$ and $(a-x)$ as denominators add up to a fraction having $(a^2 - x^2)^2$ as denominator? Source : Olry Terquem, Exercices de mathématiques élémentaires , $1842$
( At Archive.org https://archive.org/details/exercicesdemath00terqgoog/page/n85/mode/1up)
I'm working on problem  $35$ below, where appears  also the solution.

My question is simply : is this solution correct, as far as the denominator is concerned?
When I do $(a-x)^2(a+x)^2(a-x)$, I get
$ (a-x)(a-x)(a+x)(a+x)(a-x) $
$= (a-x)(a+x)(a-x)(a+x)(a-x)$
$= [(a-x)(a+x)] [(a-x)(a+x)](a-x)$
$= (a^2- x^2) ^2 (a-x)$.
Or, am I wrong?

$\require{cancel}$
We do not need the product of all denominators if one is a factor of another. For instance
$$\quad \dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{8}\\
=\dfrac{8}{24}+\dfrac{6}{24}+\dfrac{3}{24}$$
where the denominator is
$\space LCM(3,4,8)=24$ and not $3\times4\times8=96$
$$LCM \big((a-x)^2 , (a+x)^2, (a-x)\big)\\
=  (a-x)^2(a+x)^2\cancel{(a-x)}\\
=(a^2-x^2)^2$$
