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??
 A: HINT:
You could further find $$z+\frac{1}{z}=3$$
and note that $$a^3+b^3=(a+b)(a^2-ab+b^2)$$
A: Work out $a=z+1/z$. Note that $z^3+1/z^3=a^3-3a$.
A: Hint. Use the identity
$$z^{m+n}+\frac{1}{z^{m+n}}=\left( z^m+\frac{1}{z^{m}}\right)\left(z^{n}+\dfrac{1}{z^{n}}\right)-\left(z^{m-n}+\frac{1}{z^{m-n}}\right).$$
A: 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*}
A: 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
$$z^3+\frac{1}{z^3}=-3(7-1)=-18$$
A: 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 $$z^3+\frac{1}{z^3}=a^3-3a=\\(\pm 3)^3-3(\pm 3)$$
A: 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 $a=3, b=1$, 
$$S_3S_1=S_4+S_2\\
S_3=\frac {S_4+S_2}{S_1}=\frac {47+7}3=\color{red}{18}$$
