If $z^4 + \frac1{z^4}=47$ then find the value of $z^3+\frac1{z^3}$ If $z^4 + \dfrac {1}{z^4}=47$ then find the value of $z^3+\dfrac {1}{z^3}$
My Attempt:
$$z^4 + \dfrac {1}{z^4}=47$$
$$(z^2+\dfrac {1}{z^2})^2 - 2=47$$
$$(z^2 + \dfrac {1}{z^2})^2=49$$
$$z^2 + \dfrac {1}{z^2}=7$$
How do I proceed further??

Use the same trick again to get 
\begin{eqnarray*}
(z+\frac{1}{z})^2-2=7 \\
z+\frac{1}{z}=3
\end{eqnarray*}
\begin{eqnarray*}
(z+\frac{1}{z})(z^2+\frac{1}{z^2})=(z^3+\frac{1}{z^3})+(z+\frac{1}{z})=21 \\
z^3+\frac{1}{z^3}=\color{red}{18}.
\end{eqnarray*}
