\frac{x^3+1}{4-x^2} dla x=\sqrt3-1
\frac{x^3+1}{4-x^2}=\frac{(\sqrt3-1)^3+1}{4-(\sqrt3-1)^2}=
=\frac{\sqrt{27}-3(\sqrt3)^2+3\sqrt3-1+1}{4-(3-2\sqrt3+1)}=
=\frac{3\sqrt3-9+3\sqrt3}{4-4+2\sqrt3}=
=\frac{6\sqrt3-9}{2\sqrt3}*\frac{\sqrt3}{\sqrt3}=
=\frac{18-9\sqrt3}{2*3}=\frac{3(6-3\sqrt3)}{2*3}=\frac{6-3\sqrt3}{2}
Zastosowano wzory
(a-b)^2=a^2-2ab+b^2
(a-b)^3=a^3-3a^2b+3ab^2-b^3