How Did 1886 Dutch Students Solve This Complex Math Problem?

[chwala]

So what? You get one value of x for one value of a. The OP has the solution which has TWO values of x for a given value of a. The OP also has the solution for other values of a. You are missing a lot of solutions.

I will check if I can get other values by changing my $m$ value. Cheers man!

 

[Frabjous]

You are going to come up with a finite number number ‘n‘ of solutions. Given that there are an infinite number of solutions, your grade for the problem will be n/∞=0.

 

[SammyS]

@chwala ,
Your response to my post (#22), although I made a mistake in that post of mine. It is corrected here:

(My post (#22) in reference to your post (#19) )
It looks like you intended to divide both sides buy $(a+x)^\frac{1}{3}$ but you neglected dividing $5(a-x)^\frac{1}{3}$ by $(a+x)^\frac{1}{3}$ .

Correction:
... you neglected to divide $6(a-x)^\frac{2}{3}$ by $(a+x)^\frac{1}{3}$ .

Your post(#23)

This may be unorthodox approach but hey let us do it!

We have,

$(a+x)^{\frac{2}{3}} + 6 (a-x)^{\frac{2}{3}}=5(a^2-x^2)^{\frac{1}{3}}$

I will still let

$m=a+x$

and

$k=a-x$

then,

$m^{\frac{2}{3}} + 6k^{\frac{2}{3}}=5m^{\frac{1}{3}}k^{\frac{1}{3}}$

(I'll cut in here.)

This looks fine. In fact you could write that as;

$\displaystyle m^{\frac{2}{3}} -5m^{\frac{1}{3}}k^{\frac{1}{3}}+ 6k^{\frac{2}{3}}=0$ ,

which can be factored quite nicely, by the way.

You then go on to state;

$6k^{\frac{2}{3}}=m^{\frac{1}{3}}[5k^{\frac{1}{3}}-m^{\frac{1}{3}}]$

Let

$m^{\frac{1}{3}}=1$

then,

$6k^{\frac{2}{3}}-5k^{\frac{1}{3}}+1=0$
...

the other steps to solution shall remain as shown on the quoted post.

cheers!

Although you get the same quadratic equation as you did in the "quoted post" (#19), that is only due to the choice of $m=1$.

For other values of $m$, posts 19 and 22 will not agree.

 

[chwala]

@chwala ,
Your response to my post (#22), although I made a mistake in that post of mine. It is corrected here:Your post(#23)
(I'll cut in here.)

This looks fine. In fact you could write that as;

$\displaystyle m^{\frac{2}{3}} -5m^{\frac{1}{3}}k^{\frac{1}{3}}+ 6k^{\frac{2}{3}}=0$ ,

which can be factored quite nicely, by the way.

You then go on to state;

Although you get the same quadratic equation as you did in the "quoted post" (#19), that is only due to the choice of $m=1$.

For other values of $m$, posts 19 and 22 will not agree.

Thanks can see that we end up with what was previously found by other members:

... i.e

$x^2-5xy+6y^2=0$

$(x-2y)(x-3y)=0$

...

where;

$x=m^\frac{1}{3}$

and

$y=k^\frac{1}{3}$

 

[vanhees71]

Based on @vanhees71's subsequent response, he evidently meant "integration by parts." It's possible that he conflated the names of the techniques of integration by parts and integration by partial fraction decomposition.

I think it's a "Germanism", because in German we call this technique "partielle Integration". Obviously in English it's exclusively called "integration by parts".