Răspuns :
x³-x=0⇔x(x²-1)=0⇔x(x-1)(x+1)=0⇔x1=-1;x2=0;x3=1
a)x1+x2+x3=-1+0+1=0
b)(x1)³+(x2)³+(x3)³=(-1)³+0³+1³=-1+0+1=0
c) -1 0 1
0 1 -1 =0+0+0-1+1-0=0
1 -1 0
a)x1+x2+x3=-1+0+1=0
b)(x1)³+(x2)³+(x3)³=(-1)³+0³+1³=-1+0+1=0
c) -1 0 1
0 1 -1 =0+0+0-1+1-0=0
1 -1 0