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??

Since $z^2+\frac{1}{z^2}=7$, we obtain $z+\frac{1}{z}=3$ or $z+\frac{1}{z}=-3$.
In the first case we obtain
$$z^3+\frac{1}{z^3}=3\left(z^2+\frac{1}{z^2}-1\right)=3(7-1)=18$$
in the second case we obtain
$$z^3+\frac{1}{z^3}=-3(7-1)=-18$$
