[tex]\sqrt[3]{x-1} =1-x \\ (\sqrt[3]{x-1})^{3} = (1-x)^{3} \\ x-1=1-3x+3 x^{2} - x^{3} \\ x^{3} -3 x^{2} +4x-2=0 \\ \\ Conform ~teoremei~lui~Bezout~gasim~o~valoare~a~lui~x~pentru~ca \\ P(x)~sa~aiba~valoarea~0.~O~astfel~de~valoare~este~x=1. \\ Deci,~vom~imparti~polinomul~de~gradul~3~obtinut~la \\ binomul~x-1.~(impartirea~polinomului~e~in~atasament). \\ Avem: \\ (x-1)( x^{2} -2x+2)=0 \\ \\ \left \{ {{x-1=0} \atop { x^{2} -2x+2=0}} \right. \\ \left \{ {{x=1} \right. Ф[/tex]
