A Different Approach!

Using Descarte's rule of signs, we see that $x^{3}-x-1$ has 1 sign change, therefore one real solution. Trying to solve via synthetic division gives a remainder, so we need another method to solve it. We rewrite the equation with an x-squared term: $x^{3}+0x^{2}-x-1$ Since we cannot solve through factoring: $x=\sqrt[3]{\frac{1}{2}+\sqrt{(\frac{-1}{3})^{3}+(\frac{1}{2})^{2}}}\hspace{2mm}+\sqrt[3]{\frac{1}{2}-\sqrt{(\frac{-1}{3})^{3}+(\frac{1}{2})^{2}}}\hspace{2mm}-(\frac{1}{3})\cdot 0\approx 1.65$ Now we add: $\frac{1}{2.65}+\frac{1}{1+0}+\frac{1}{1+0}\approx 2.3$ So we round down, and get a result of 2.

a+b+c=0 ab+bc+ac=-1 abc=1 Putting above values in simplified expression we get 2