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?

Question

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?

Answer 1

Your calculations are correct, but you slip up when it comes to what you actually do to the denominator to add fractions.

Specifically, when you multiply each addand fraction by a form of $1$ to get get a common denominator, that denominator is not (necessarily) the product of the individual denominators, but their least common multiple.

That noted, it should be clear that

$${\rm lcm}((a-x)^2,(a+x)^2,a-x)=(a-x)^2(a+x)^2=(a^2-x^2)^2$$

Answer 2

$\require{cancel}$

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 $3\times4\times8=96$

$$LCM \big((a-x)^2 , (a+x)^2, (a-x)\big)\\ = (a-x)^2(a+x)^2\cancel{(a-x)}\\ =(a^2-x^2)^2$$