Multiply both sides of the first equation by [tex]2^x[/tex]:
[tex]2^x-2^{-x}=4\implies 2^{2x}-1=4\cdot2^x\implies 2^{2x}-4\cdot2^x-1=0[/tex]
This is quadratic in [tex]2^x[/tex]; to make this clear, substitute [tex]y=2^x[/tex]. Then
[tex]y^2-4y-1=0\implies y=2\pm\sqrt5[/tex]
One of these solutions for [tex]y[/tex] is negative. But, if [tex]x[/tex] is real, then [tex]y=2^x[/tex] is always supposed to be positive, so we can throw out the negative root, leaving [tex]y=2+\sqrt5[/tex].
We actually don't have to solve for [tex]x[/tex] exactly. We can just rewrite the next two equations in terms of [tex]y[/tex].
[tex]2^{2x}+2^{-2x}=(2^x)^2+(2^{-x})^2=y^2+\dfrac1{y^2}[/tex]
[tex]2^{3x}-2^{-3x}=y^3-\dfrac1{y^3}[/tex]
Since [tex]y=2+\sqrt5[/tex], we get
[tex](2+\sqrt5)^2+\dfrac1{(2+\sqrt5)^2}=18[/tex]
[tex](2+\sqrt5)^3-\dfrac1{(2+\sqrt5)^3}=76[/tex]
