<s>[INST] What is $x^3 + x + 1$ for $x=\frac{\sqrt 5 - 1}{2}$ 
What is $x^3 + x + 1$ for $x=\frac{\sqrt 5 - 1}{2}$

Of course we can subtitute and expand it
${(\frac{\sqrt 5 -1}{2})}^3 + \frac{\sqrt 5 -1}{2} + 1$
Is there a better way to do it?
 [/INST] Yes. $\dfrac{\sqrt{5}-1}{2}$ is a root of $x^2+x-1=0$. So $x^2=1-x$.
Multiply by $x$ to get 
$$x^3=x-x^2=x-1+x=2x-1$$
Therefore 
$$x^3+x+1=3x=\frac{3}{2}(\sqrt{5}-1)$$
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