Докажите тождество (b+1)^3-3(b+1)^2+3(b+1)-1=b^3 a^4+5a^3+9a^2+7a+2=(a+1)^3(a+2)

[latex](b+1)^3-3(b+1)^2+3(b+1)-1=b^3[/latex] [latex]b^3+3b^2+3b+1-3(b^2+2b+1)+3(b+1)-1=b^3[/latex]  [latex]b^3+3b^2+3b+1-3b^2-6b-3+3b+3-1=b^3[/latex]  [latex]b^3+3b^2+3b+1-3b^2-6b-3+3b+3-1=b^3[/latex]  [latex]b^3=b^3[/latex]    [latex]a^4+5a^3+9a^2+7a+2=(a+1)^3(a+2)[/latex] [latex]a^4+5a^3+9a^2+7a+2=(a^3+3a^2+3a+1)(a+2)[/latex] [latex]a^4+5a^3+9a^2+7a+2= [/latex]  [latex]=a^4+3a^3+3a^2+a+2a^3+6a^2+6a+2[/latex]   [latex]a^4+5a^3+9a^2+7a+2=a^4+5a^3+9a^2+7a+2 [/latex]