Solution #8

Let a = 2^x - 4, and let b = 4^x - 2. Then our equation can be rewritten as:

a³ + b³ = (a+b)³

Expanding the right-hand side, we have:

a³ + b³ = a³ + 3a²b + 3ab² + b³

And hence:

3a²b + 3ab² = 0
ab(a+b) = 0

We thus have three cases: a = 0, b = 0, a = -b.

If a = 0, then 2^x = 4, and x = 2.

If b = 0, then 4^x = 2, and x = 1/2. If a = -b, then 2^x - 4 = 2 - 4^x, or:

4^x + 2^x - 6 = 0

4^x = (2^x)², so this factors to:

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

So 2^x = -3, or 2^x = 2. But 2^x is never negative, so we ignore that solution. 2^x yields x = 1.

The three solutions are thus 2, 1, and 1/2, and their sum is 7/2.