Page:Laplace - Œuvres complètes, Gauthier-Villars, 1878, tome 8.djvu/63

Supposons qu’une des racines de $x$ soit $p-f,$ et nommant, pour abréger, $\mathrm {K}$ le second membre de l’équation $(\omega ),$ nous aurons

${\frac {1}{(2p+2q+f)^{5}}}-{\frac {1}{3(3p+2q+f)^{5}}}-{\frac {1}{3(2p+3q+f)^{5}}}=\mathrm {K} \,;$

supposons que $p-f-u$ soit une seconde racine de $x,\ f+u$ étant positif et moindre que $p\,;$ nous aurons

${\begin{aligned}&{\frac {1}{(2p+2q+f)^{5}\left(1+{\frac {u}{2p+2q+f}}\right)^{5}}}\\-&{\frac {1}{3(3p+2q+f)^{5}\left(1+{\frac {u}{3p+2q+f}}\right)^{5}}}\\-&{\frac {1}{3(2p+3q+f)^{5}\left(1+{\frac {u}{2p+3q+f}}\right)^{5}}}=\mathrm {K} .\\\end{aligned}}$

Soit

${\begin{aligned}&{\frac {1}{(2p+2q+f)^{5}\left(1+{\frac {u}{2p+2q+f}}\right)^{5}}}={\frac {1}{(2p+2q+f)^{5}}}\left(1+{\frac {1}{l}}\right),\\-&{\frac {1}{3(3p+2q+f)^{5}\left(1+{\frac {u}{3p+2q+f}}\right)^{5}}}={\frac {1}{(3p+2q+f)^{5}}}\left(1+{\frac {1}{l'}}\right),\\-&{\frac {1}{3(2p+3q+f)^{5}\left(1+{\frac {u}{2p+3q+f}}\right)^{5}}}={\frac {1}{(2p+3q+f)^{5}}}\left(1+{\frac {1}{l''}}\right)\,;\end{aligned}}$

$l,l'$ et $l''$ seront positifs ou négatifs, suivant que $u$ sera positif ou négatif ; de plus, on aura $l<l'$ et $l<l'',$ ensuite on aura

${\frac {1}{l(2p+2q+f)^{5}}}-{\frac {1}{3l'(3p+2q+f)^{5}}}-{\frac {1}{3l''(2p+3q+f)^{5}}}=0\,;$

mais on a

${\frac {1}{l(2p+2q+f)^{5}}}-{\frac {1}{3l(3p+2q+f)^{5}}}-{\frac {1}{3l(2p+3q+f)^{5}}}={\frac {\mathrm {K} }{l}}\,;$

donc

${\frac {\mathrm {K} }{l}}+{\frac {1}{3(3p+2q+f)^{5}}}\left({\frac {1}{l}}-{\frac {1}{l'}}\right)+{\frac {1}{3(2p+3q+f)^{5}}}\left({\frac {1}{l}}-{\frac {1}{l'}}\right)=0.$