Equation

$Evaluate\\ a^3- \dfrac{1}{a^3}\\ if\\ a^2 - a - 1 = 0.\\ a=\frac{1}{2}\pm\sqrt{\frac{1}{4}+1}=\frac{1}{2}\pm\sqrt{\frac{1+4}{4}}=\frac{1}{2}(1\pm \sqrt{5})\\ \color{blue}a\in\{1.618, -0.618\}$

$a^3- \dfrac{1}{a^3}\\ =\dfrac{a^6-1}{a^3}\\ =\dfrac{\dfrac{1}{64}(1\pm\sqrt{5})^6-1}{\dfrac{1}{8}(1\pm \sqrt{5})^3}\\ =\dfrac{1}{8}(1\pm\sqrt{5})^3-\dfrac{1}{\dfrac{1}{8}(1\pm \sqrt{5})^3}\\ \color{blue} a^3- \dfrac{1}{a^3}=4$