How to simplify ${(1+2i)}^6$? How to simplify ${(1+2i)}^6$ using De Moivre's formula?
I have found that $r=\sqrt 5$ and $\tan x=2$ but I can't find the exact value of $x$.
Hint You have that $\tan(\theta) = 2$ and you need $\cos(6\theta)$ and $\sin(6\theta).$ Drawing a triangle in the first quadrant you can see that if $\tan(\theta) = 2$ you have $\cos(\theta) = 1/\sqrt{5}$ and $\sin(\theta) = 2/\sqrt{5}.$ Now can you find $\sin(6\theta)$ and $\cos(6\theta)$?

As others have said, it might be easier to use a hybrid approach where you square or cube first and then do the rest with trig.
Using De Moivre for this computation doesn't make much sense.

Either you use it numerically and evaluate $\sqrt5^6e^{i6\arctan2}=125(0.936+i0.352)$, or you establish the analytical formulas for $\cos6x,\sin6x$. In both cases, the computation is more sophisticated than direct evaluation by the binomial formula

$$(1+i2)^6=1+i6\cdot2-15\cdot4-i20\cdot8+15\cdot16+i6\cdot32-64=117+i44$$
or, faster,
$$(1+i2)^6=(-3+4i)^3=-3^3+i3\cdot3^2\cdot4+3\cdot3\cdot4^2-i4^3=117+i44.$$