<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] Note that :suppose $a=z+\frac 1z $so 
$$z+\frac 1z=a\\z^2+\frac{1}{z^2}=(z+\frac 1z)^2-2z\frac1z=a^2-2\\
z^4+\frac{1}{z^4}=(z^2+\frac{1}{z^2})^2-2=(a^2-2)^2-2\\
z^3+\frac{1}{z^3}=(z+\frac 1z)^3-3z\frac1z(z+\frac 1z)=a^3-3a$$
what you have now is $$(a^2-2)^2-2=47 \to\\(a^2-2)^2=49 \\\begin{cases}a^2-2=7 & a^2=9\\\to a^2-2=-7 & a^2=-5 \end{cases}
\\a^2-2= 7 \\\to a^2=9 \\\to a=\pm 3 $$now 
you want $$z^3+\frac{1}{z^3}=a^3-3a=\\(\pm 3)^3-3(\pm 3)$$
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