dezvoltand, dupa formula lui (a+b)³= a³+3a²b+3ab²+b³
pe (1+√2)³ obtinem 1+3√2+3* (√2)² +2√2=7+5√2
si pe (1-√2)³=1-3*1*√2+3*1*2-2√2=7-5√2
deci
x= ∛(1+√2)³ =1+√2
y=∛(1-√2)³=1-√2
1+√2=a+√2⇒a=1
x=y=1+√2+1-√2=2∈Q adica ∉R\Q
A3 . analog
∛(1+√5)³- ∛(1-√5)³=1+√5-(1-√5)=√5+√5=2√5
∛[1/2+√(71/12)]³-∛[√(71/12)-1/2]³=1/2-√(71/12)-√(71/12)+1/2=1/2+1/2=1
