Dans ce parcours, il faut utiliser le schéma suivant :
Soit 
Pour l’utilisation de ce dessin, se reporter aux conditions exposées dans
» identité du mois de Janvier 2022 »
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Soit x =
![\sqrt[3]{(n+1)^2}+\sqrt[3]{n(n+1)}+\sqrt[3]{n^2}\\\](http://s0.wp.com/latex.php?latex=%5Csqrt%5B3%5D%7B%28n%2B1%29%5E2%7D%2B%5Csqrt%5B3%5D%7Bn%28n%2B1%29%7D%2B%5Csqrt%5B3%5D%7Bn%5E2%7D%5C%5C%5C&bg=T&fg=000000&s=0 "\sqrt[3]{(n+1)^2}+\sqrt[3]{n(n+1)}+\sqrt[3]{n^2}\\\")
Alors ![[x-\frac{1}{x^2}]^3\](http://s0.wp.com/latex.php?latex=%5Bx-%5Cfrac%7B1%7D%7Bx%5E2%7D%5D%5E3%5C&bg=T&fg=000000&s=0 "[x-\frac{1}{x^2}]^3\")
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Soit x =
![\sqrt[6]{4}+\sqrt[6]{2}+\sqrt[6]{1}\\\](http://s0.wp.com/latex.php?latex=%5Csqrt%5B6%5D%7B4%7D%2B%5Csqrt%5B6%5D%7B2%7D%2B%5Csqrt%5B6%5D%7B1%7D%5C%5C%5C&bg=T&fg=000000&s=0 "\sqrt[6]{4}+\sqrt[6]{2}+\sqrt[6]{1}\\\")
Alors
![x^3-3 = (3x^2)\sqrt[6]{2}-2\sqrt{2}\\](http://s0.wp.com/latex.php?latex=x%5E3-3+%3D+%283x%5E2%29%5Csqrt%5B6%5D%7B2%7D-2%5Csqrt%7B2%7D%5C%5C&bg=T&fg=000000&s=0 "x^3-3 = (3x^2)\sqrt[6]{2}-2\sqrt{2}\\")
Soit x =
![\sqrt[6]{9}+\sqrt[6]{6}+\sqrt[6]{4}\\\](http://s0.wp.com/latex.php?latex=%5Csqrt%5B6%5D%7B9%7D%2B%5Csqrt%5B6%5D%7B6%7D%2B%5Csqrt%5B6%5D%7B4%7D%5C%5C%5C&bg=T&fg=000000&s=0 "\sqrt[6]{9}+\sqrt[6]{6}+\sqrt[6]{4}\\\")
Alors
![x^3-5 = (3x^2)\sqrt[6]{6}-2\sqrt{6}\\](http://s0.wp.com/latex.php?latex=x%5E3-5+%3D+%283x%5E2%29%5Csqrt%5B6%5D%7B6%7D-2%5Csqrt%7B6%7D%5C%5C&bg=T&fg=000000&s=0 "x^3-5 = (3x^2)\sqrt[6]{6}-2\sqrt{6}\\")
Soit x =
![\sqrt[6]{16}+\sqrt[6]{12}+\sqrt[6]{9}\\](http://s0.wp.com/latex.php?latex=%5Csqrt%5B6%5D%7B16%7D%2B%5Csqrt%5B6%5D%7B12%7D%2B%5Csqrt%5B6%5D%7B9%7D%5C%5C&bg=T&fg=000000&s=0 "\sqrt[6]{16}+\sqrt[6]{12}+\sqrt[6]{9}\\")
Alors
![x^3-7 = (3x^2)\sqrt[6]{12}-2\sqrt{12}\\](http://s0.wp.com/latex.php?latex=x%5E3-7+%3D+%283x%5E2%29%5Csqrt%5B6%5D%7B12%7D-2%5Csqrt%7B12%7D%5C%5C&bg=T&fg=000000&s=0 "x^3-7 = (3x^2)\sqrt[6]{12}-2\sqrt{12}\\")
Soit x =
![\sqrt[6]{25}+\sqrt[6]{20}+\sqrt[6]{16}\\\](http://s0.wp.com/latex.php?latex=%5Csqrt%5B6%5D%7B25%7D%2B%5Csqrt%5B6%5D%7B20%7D%2B%5Csqrt%5B6%5D%7B16%7D%5C%5C%5C&bg=T&fg=000000&s=0 "\sqrt[6]{25}+\sqrt[6]{20}+\sqrt[6]{16}\\\")
Alors
![x^3-9 = (3x^2)\sqrt[6]{20}-2\sqrt{20}\\](http://s0.wp.com/latex.php?latex=x%5E3-9+%3D+%283x%5E2%29%5Csqrt%5B6%5D%7B20%7D-2%5Csqrt%7B20%7D%5C%5C&bg=T&fg=000000&s=0 "x^3-9 = (3x^2)\sqrt[6]{20}-2\sqrt{20}\\")
Soit x =
![\sqrt[6]{36}+\sqrt[6]{30}+\sqrt[6]{25}\\\](http://s0.wp.com/latex.php?latex=%5Csqrt%5B6%5D%7B36%7D%2B%5Csqrt%5B6%5D%7B30%7D%2B%5Csqrt%5B6%5D%7B25%7D%5C%5C%5C&bg=T&fg=000000&s=0 "\sqrt[6]{36}+\sqrt[6]{30}+\sqrt[6]{25}\\\")
Alors
![x^3-11 = (3x^2)\sqrt[6]{30}-2\sqrt{30}\\](http://s0.wp.com/latex.php?latex=x%5E3-11+%3D+%283x%5E2%29%5Csqrt%5B6%5D%7B30%7D-2%5Csqrt%7B30%7D%5C%5C&bg=T&fg=000000&s=0 "x^3-11 = (3x^2)\sqrt[6]{30}-2\sqrt{30}\\")
…
généralisation
Soit x =
![\sqrt[6]{(n+1)^2}+\sqrt[6]{n(n+1)}+\sqrt[6]{n^2}\\\](http://s0.wp.com/latex.php?latex=%5Csqrt%5B6%5D%7B%28n%2B1%29%5E2%7D%2B%5Csqrt%5B6%5D%7Bn%28n%2B1%29%7D%2B%5Csqrt%5B6%5D%7Bn%5E2%7D%5C%5C%5C&bg=T&fg=000000&s=0 "\sqrt[6]{(n+1)^2}+\sqrt[6]{n(n+1)}+\sqrt[6]{n^2}\\\")
Alors
![x^3-(2n+1) = (3x^2)\sqrt[6]{n(n+1)}-2\sqrt{n(n+1)}\\](http://s0.wp.com/latex.php?latex=x%5E3-%282n%2B1%29+%3D+%283x%5E2%29%5Csqrt%5B6%5D%7Bn%28n%2B1%29%7D-2%5Csqrt%7Bn%28n%2B1%29%7D%5C%5C&bg=T&fg=000000&s=0 "x^3-(2n+1) = (3x^2)\sqrt[6]{n(n+1)}-2\sqrt{n(n+1)}\\")
n prend la suite des nombres entiers
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Soit x =
![\sqrt[9]{(n+1)^2}+\sqrt[9]{n(n+1)}+\sqrt[9]{n^2}\\\](http://s0.wp.com/latex.php?latex=%5Csqrt%5B9%5D%7B%28n%2B1%29%5E2%7D%2B%5Csqrt%5B9%5D%7Bn%28n%2B1%29%7D%2B%5Csqrt%5B9%5D%7Bn%5E2%7D%5C%5C%5C&bg=T&fg=000000&s=0 "\sqrt[9]{(n+1)^2}+\sqrt[9]{n(n+1)}+\sqrt[9]{n^2}\\\")
Alors
![x^3 = (3x^2)\sqrt[9]{n(n+1)}+[\sqrt[3]{n+1}-\sqrt[3]{n}]^2\\\](http://s0.wp.com/latex.php?latex=x%5E3+%3D+%283x%5E2%29%5Csqrt%5B9%5D%7Bn%28n%2B1%29%7D%2B%5B%5Csqrt%5B3%5D%7Bn%2B1%7D-%5Csqrt%5B3%5D%7Bn%7D%5D%5E2%5C%5C%5C&bg=T&fg=000000&s=0 "x^3 = (3x^2)\sqrt[9]{n(n+1)}+[\sqrt[3]{n+1}-\sqrt[3]{n}]^2\\\")
