1(a+b)³=a³+3a²b+3ab²+b³= a³+b³+3ab(a+b)=a³+b³+3ab=1
2a2=a+1 a^2-a=1
a5-5a+2=a(a^4-5)+2=a[(a+1)^2-5]+2=a(a^2+2a-4)+2=a(a+1+2a-4)+2=a*(3a-3)+2=3(a^-a)+2=3+2=5
1(a+b)³=a³+3a²b+3ab²+b³= a³+b³+3ab(a+b)=a³+b³+3ab=1
2a2=a+1 a^2-a=1
a5-5a+2=a(a^4-5)+2=a[(a+1)^2-5]+2=a(a^2+2a-4)+2=a(a+1+2a-4)+2=a*(3a-3)+2=3(a^-a)+2=3+2=5