đặt $x^{3}$ -2x +3 =a, ta có:
<=>(x^3-3x+3)(x^3-2x+3)=2x^2
<=> (a-x)a=2x^2
<=>$a^{2}$ -ax -2$x^{2}$ =0
<=>($a^{2}$ - $x^{2}$ )- (ax +$x^{2}$ ) =0
<=>(a+x)(a-x) - x(a+x) =0
<=>(a+x)(a-2x)=0
=>\(\left[ \begin{array}{l}a+x=0(1)\\a+2x=0(2)\end{array} \right.\)
(1)<=> $x^{3}$ -2x +3 +x=0 =>$x^{3}$ -x +3=0
(2)<=>$x^{3}$ -2x +3 -2x =0 =>$x^{3}$ -4x +3 =0
=>($x^{3}$ -X) -(3X-3)=0
=>x($x^{2}$ -1) -3(x-1)=0
=>(x-1)(x.(x+1)-3)=0
=>\(\left[ \begin{array}{l}x=1\\x^{2}+ x -3=0 \end{array} \right.\)
=>\(\left[ \begin{array}{l}x=1\\\left[ \begin{array}{l}\frac{x=-1 +\sqrt[]{13}}{2} \\\frac{x=-1-\sqrt[]{13}}{2} \end{array} \right.\end{array} \right.\)
