Упростить и решить [latex] \frac{a^{3}+ b^{3}+3b^{2}+3b+1}{ a^{2}-ab-a+(b+1)^2 } a= -3- \sqrt[5]{3} ,b=11+ \sqrt[5]{3} [/latex]

[latex]\frac{a^{3}+b^{3}+3b^{2}+3b+1}{a^{2}-ab-a+(b+1)^2}=\frac{a^{3}+(b+1)^3}{a^{2}-ab-a+(b+1)^2}=\\\ =\frac{(a+b+1)(a^{2}-ab-a+(b+1)^2)}{a^{2}-ab-a+(b+1)^2}=a+b+1\\\ \\\ \ \ \ a= -3- \sqrt[5]{3} \ \ \ \ b=11+ \sqrt[5]{3} \\\ \\\ -3- \sqrt[5]{3}+11+ \sqrt[5]{3}+1=9[/latex]