Algebra

\({a}^{3}-\frac{1}{{a}^{3}}=(a-\frac{1}{a})({a}^{2}+1+\frac{1}{{a}^{2}})=(a-\frac{1}{a})({(a-\frac{1}{a})}^{2}+3)\), which is what we want to find.

We see \(a-\frac{1}{a}\) will me good to know.

We have:

\({a}^{2}-a-1=0\), divide both sides by a.

\(a-1-\frac{1}{a}=0\), \(a-\frac{1}{a}=1\).

Plugging in, \({a}^{3}-\frac{1}{{a}^{3}}=(1)({1}^{2}+3)=4\).

So, \(\underline{{a}^{3}-\frac{1}{{a}^{3}}}\) is 4.