How to quickly solve a series of exponential equations. 
When $2^x - 2^{-x} = 4$, then  $2^{2x} + 2^{-2x} =\hbox{ ?}$ and $2^{3x} - 2^{-3x} =\hbox{ ?}$

I have a doubt with this one. I assume I should use some kind of formula to solve it because if I solve the first equation I get that $x = \log_2(1 + \sqrt5)$ and then I just solve the other two equations and get that the result of the second one is $18$ and the last one is $76$.
The thing is that this exercise is supposed to be done quickly but this method takes a lot of time to calculate. Is there another way to solve it that I'm missing?
 A: hint
Let $$X=2^x$$
then
$$2^x-2^{-x}=X-\frac 1X=4$$
$$(X-\frac 1X)^2=X^2+\frac{1}{X^2}-2$$
$$=4^2=16$$
thus
$$2^{2x}+2^{-2x}=X^2+\frac{1}{X^2}=16+2=18$$
Now do the same with $(X-\frac {1}{X})^3$.
A: Denote $2^x=:t$, thus $t-\frac{1}{t}=4$, then
$$16=\left(t-\frac{1}{t}\right)^2=t^2+\frac{1}{t^2}-2
\Rightarrow t^2+\frac{1}{t^2}=18$$
$$4\cdot 18=\left(t-\frac{1}{t}\right)
\left(t^2+\frac{1}{t^2}\right)=t^3-\frac{1}{t^3}-t+\frac{1}{t}=t^3-\frac{1}{t^3}-4\Rightarrow$$
$$t^3-\frac{1}{t^3}=4\cdot 18+4=76$$
A: \begin{align}
2^x-2^{-x}&=4\\
2^x-4-2^{-x}&=0\\
2^{2x}-4\cdot2^x-1&=0\\
2^x&=2+\sqrt5\\
\end{align}
As such, we have $$2^{2x}+2^{-2x}=(2+\sqrt5)^2+\frac1{(2+\sqrt5)^2}$$ and $$2^{3x}-2^{-3x}=(2+\sqrt5)^3-\frac1{(2+\sqrt5)^3}.$$
A: From $2^x-2^{-x}=4$ you get $(2^x-2^{-x})^2=16$; by squaring,
$$
2^{2x}-2\cdot2^x\cdot2^{-x}+2^{-2x}=16
$$
that yields
$$
2^{2x}+2^{-2x}=16+2=18
$$
Finally
$$
2^{3x}-2^{-3x}=(2^x-2^{-2x})(2^{2x}+2^x\cdot2^{-x}+2^{-2x})=4\cdot(18+1)=76
$$
from $a^3-b^3=(a-b)(a^2+ab+b^2)$.
