How to Solve : $ A =\frac{1}{6}\left((\log_2(3))^3-(\log_2(6))^3-(\log_2(12))^3+(\log_2(24))^3\right) $ $ A =\frac{1}{6}\left((\log_2(3))^3-(\log_2(6))^3-(\log_2(12))^3+(\log_2(24))^3\right).$
Solve for $2^A.$
(no calculators or graphs are permitted)
The way I went about solving this problem was using the properties of base $2$. For example, $2^x = 8$ , $x$ is obviously $3$, and $2^y = 16$, $y = 4 = x+1$, so adding $1$ to the exponent will give twice the answer. Since the arguments are twice each other, I used this pattern: 
$ A =\frac{1}{6}\left((x)^3-(x+1)^3-(x+2)^3+(x+3)^3\right)$
$= \frac{1}{6}\left((x^3-(x^3 + 3x^2 + 3x + 1)-(x^3 + 6x^2 + 12x + 8)+(x^3 + 9x^2 + 27x + 27)\right)$
Simplifying this we get $A = (12x + 18)/6 = 2x + 3.$
So now I solved for $2^{2x+3}$ using $2^{2x} =2^x * 2^x = 3^2 = 9.$ Now, since we add $3$ to the exponent, we have $2^{2x+3} = (2^3)(9) = 72.$
Would this be the correct answer, and if not, what would be a good way of going about it?

If it were to be asked in the exam, you could save some time in the simplification exploiting the fact that $a^3-b^3 = (a-b).(a^2+ab+b^2)$
Rearanging in the following  way
$$\frac{1}{6}\left((x+3)^3 - (x+2)^3 - \left((x+1)^3 - x^3)\right)\right)$$
$$\frac{1}{6}\left((x+3)^2 + (x+2)^2+(x+3)(x+2) - \left((x+1)^2 + x^2+x(x+1))\right)\right)$$
$$\frac{1}{6}\left((x^2+6x+9+x^2+4x+4+x^2+5x+6) - \left((x^2+2x+1+x^2+x^2+x)\right)\right)$$
$$\frac{1}{6}\left((3x^2+15x+19) - \left((3x^2+3x+1)\right)\right)$$
$$\frac{1}{6}\left(12x+18\right)$$
