Here comes the boom!!!

Take $2^{x} - 4$ as $a$ , $4^{x} - 2$ as $b$ , and apply $a^{3} + b^{3} = {(a + b)}^{3}$ , which gives $a + b = 0$ and $ab = 0$ . Solving, we get $x = 2 , \frac{1}{2} , 1$ . Summing up, we get $x = \boxed{\frac{7}{2}}$ .

We know that $(2^x-4)+(4^x-2)=4^x+2^x-6$

So we can rewrite this question as $a^3+b^3=(a+b)^3$

By expanding and simplifying, we find that $0=a^2b+b^2a=ab(a+b)$

By re-substituting a and b, we find that $(2^x-4)(4^x-2)(4^x+2^x-6)=0$

If any individual multiplier is 0, this statement will be true so

$2^x-4=0$

$4^x-2=0$

$4^x+2^x-6=0$

The first two are fairly easy to simplify and we find that $\boxed{x=2, \frac{1}{2}}$

The third is not as clear to see, but it can be rewritten as a "quadratic" and factored as $(2^x)^2+2^x-6=0$

$(2^x+3)(2^x-2)=0$

Now we can see our final x value of $\boxed{x=1}$

Overall $\boxed{Sum=3.5}$