en général,
![{\displaystyle {\frac {1}{p^{m}}}+{\frac {1}{q^{m}}}+{\frac {1}{r^{m}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/82afd46c0b48583957169d04738b75d9ff92d390)
![{\displaystyle {\begin{aligned}=&\left({\frac {b}{a}}\right)^{m}-{\frac {mc}{a}}\left({\frac {b}{a}}\right)^{m-2}+{\frac {md}{a}}\left({\frac {b}{a}}\right)^{m-3}\\&+{\frac {m(m-3)}{2a^{2}}}c^{2}\left({\frac {b}{a}}\right)^{m-4}+{\frac {m(m-4)}{2a^{2}}}2cd\left({\frac {b}{a}}\right)^{m-5}+{\frac {m(m-5)}{2a^{2}}}d^{2}\left({\frac {b}{a}}\right)^{m-6}\\&-{\frac {m(m-4)(m-5)}{2.3.a^{3}}}c^{3}\left({\frac {b}{a}}\right)^{m-6}+{\frac {m(m-5)(m-6)}{2.3.a^{3}}}3c^{2}d\left({\frac {b}{a}}\right)^{m-7}\\&-{\frac {m(m-6)(m-7)}{2.3.a^{3}}}3cd^{2}\left({\frac {b}{a}}\right)^{m-8}+{\frac {m(m-7)(m-8)}{2.3.a^{3}}}d^{3}\left({\frac {b}{a}}\right)^{m-9}-\ldots \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f8a2d82ba963f7dabee80569774614ab95efa659)
cette série étant continuée jusqu’à ce qu’on parvienne à des puissances négatives de\frac{b}{a}.
6. Exemple III. — Soit l’équation
![{\displaystyle a-bx-x^{n}=0\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f589552d24b4979ce705c3fe1e8cdea34941debb)
on aura
![{\displaystyle \mathrm {X} ={\frac {x^{n}}{a}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee3511f6d0c1403ef7dcabe94d666fbb4c3f4903)
donc
![{\displaystyle \mathrm {X} ^{2}={\frac {x^{2n}}{a^{2}}},\quad \mathrm {X} ^{3}={\frac {x^{3n}}{a^{3}}},\ldots .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/587889cb8f824ab7882fde99620941652597a8f2)
donc
![{\displaystyle \alpha _{n-1}={\frac {1}{a}},\quad \beta _{2n-1}={\frac {1}{a^{2}}},\ldots ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d65d038959158f64e44aa39159c177d84a3c02cb)
et toutes les autres quantités nulles ; donc
![{\displaystyle {\begin{aligned}{\frac {1}{p^{m}}}+{\frac {1}{q^{m}}}+{\frac {1}{r^{m}}}=&\left({\frac {b}{a}}\right)^{m}\\&+{\frac {m}{a}}\left({\frac {b}{a}}\right)^{m-n}+{\frac {m(m-2n+1)}{2a^{2}}}\left({\frac {b}{a}}\right)^{m-2n}\\&+{\frac {m(m-3n+2)(m-3n+1)}{2.3.a^{3}}}c^{2}\left({\frac {b}{a}}\right)^{m-3n}\\&+{\frac {m(m-4n+3)(m-4n+2)(m-4n+1)}{2.3.4.a^{4}}}2cd\left({\frac {b}{a}}\right)^{m-4n}+\ldots \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9282ab2cde701b5f1ae48c5cf5286fd6964e37d7)
jusqu’à ce qu’on arrive à des puissances négatives de ![{\displaystyle {\frac {b}{a}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0ce05674bdb462f449549f54de5a49f2a4b8edc2)