<s>[INST] 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??
 [/INST] Try this notation for less clutter:
Let $$S_n=z^n+\frac 1{z^n}$$
It can be easily shown that
$$S_n^2=S_{2n}+2$$
Hence 
$$S_4=S_2^2-2=47
\qquad \Rightarrow S_2=7\\
S_2=S_1^2-2=7\qquad \Rightarrow S_1=3$$
Also, $$S_aS_b=S_{a+b}S_{a-b}$$
Putting $a=3, b=1$, 
$$S_3S_1=S_4+S_2\\
S_3=\frac {S_4+S_2}{S_1}=\frac {47+7}3=\color{red}{18}$$
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