Răspuns :
Folosind punctul b) => xox= (x-3)(x-3)+3= (x-3)²+3
(x o x) o x= [(x-3)²+3] o x=[(x-3)²+3-3](x-3)+3= (x-3)³+3
(x o x) o x=3 <=>
(x-3)³+3=3 <=>
(x-3)³=0 <=>
x-3=0
x=3 ∈ R
(x o x) o x= [(x-3)²+3] o x=[(x-3)²+3-3](x-3)+3= (x-3)³+3
(x o x) o x=3 <=>
(x-3)³+3=3 <=>
(x-3)³=0 <=>
x-3=0
x=3 ∈ R