<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] Hint. Use the identity
$$z^{m+n}+\frac{1}{z^{m+n}}=\left( z^m+\frac{1}{z^{m}}\right)\left(z^{n}+\dfrac{1}{z^{n}}\right)-\left(z^{m-n}+\frac{1}{z^{m-n}}\right).$$
</s>