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By x3 I assume that you mean x3. In which case f(x)=x3-2x+1, and f'(x)=3x2-2.
Therefore our iteration formula is:
xn+1=xn- (xn3-2xn+1)/(3xn2-2)
Starting with x0=0 we get:
x1=0.5
x2=0.6
x3=0.617391304
x4=0.618033095
x5=0.618033988
x6=0.618033988
Starting with x0=0.9 we get:
x1=1.065116279
x2=1.009457333
x3=1.000255451
x4=1.000000195
x5=1
x6=1
Starting with x0=-1.5 we get:
x1=-1.631578947
x2=-1.618183589
x3=-1.618034007
x4=-1.618033989
x5=-1.618033989
The 3 real roots to f(x) are x=-1.618033989, x=0.618033988, and x=1
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