x3+(x3(x−1)3+3x2x−1)−2=0⇔(x2−2x+2)(x4−x3+2x2−2x+1)(x−1)3=0
⇔x2−2x+2=0hoặc:x4−x3+2x2−2x+1=0
mà :x^2-2x+2=(x-1)^2+1>0(vô no)
con:x^4-x^3+2x^2-2x+1= 4(x^4-x^3+\frac{x^2}{4})+7x^2-8x+4=4(x^2-\frac{x}{2})^2+(x-\frac{4}{7})^2+\frac{12}{7}>0
tóm lại pt vô no