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.