help

add 2, take a sqrt cube it and then expand

If
\(a^2 + \dfrac{1}{a^2} \)= 1,

then find
 \(a^3 + \dfrac{1}{a^3}\).

\(\begin{array}{|rcll|} \hline \left( a^2 + \dfrac{1}{a^2} \right) \left( a+\dfrac{1}{a} \right) &=& a^3+\dfrac{a^2}{a}+\dfrac{a}{a^2} + \dfrac{1}{a^3} \\\\ \left( a^2 + \dfrac{1}{a^2} \right) \left( a+\dfrac{1}{a} \right) &=& a^3+ \dfrac{1}{a^3}+ a +\dfrac{1}{a} \\\\ \left( a^2 + \dfrac{1}{a^2} \right) \left( a+\dfrac{1}{a} \right)-\left( a+\dfrac{1}{a} \right) &=& a^3+ \dfrac{1}{a^3} \\\\ \left( a+\dfrac{1}{a} \right) \left( a^2 + \dfrac{1}{a^2} -1 \right) &=& a^3+ \dfrac{1}{a^3} \quad | \quad a^2 + \dfrac{1}{a^2} = 1 \\\\ \left( a+\dfrac{1}{a} \right) \left(1 -1 \right) &=& a^3+ \dfrac{1}{a^3} \\\\ \left( a+\dfrac{1}{a} \right) \times 0 &=& a^3+ \dfrac{1}{a^3} \\\\ 0 &=& a^3+ \dfrac{1}{a^3} \\\\ \mathbf{ a^3+ \dfrac{1}{a^3} } &=& \mathbf{0} \\ \hline \end{array}\)