RĂ©ponse :
On a [tex]a^{2}+ b^{2}+ c^{2} =1[/tex]
Alors, on calcule :
[tex]a^{4} + (ab+c)^{2} + (ac-b)^{2}\\=a^{4} + a^{2} b^{2} + 2abc + c^{2} + a^{2} c^{2} - 2acb +b^{2} \\= a^{4} + a^{2} b^{2} + c^{2} + a^{2} c^{2}+ b^{2} \\= a^{2} (b^{2} +c^{2} + a^{2} ) + c^{2} + b^{2} \\= a^{2} (1)+ c^{2} + b^{2}\\=a^{2} + c^{2} + b^{2}\\= 1[/tex]
