We first cube each side of the equation to get $x^2 - 4x + 4 = 16^3$.  Notice that $x^2 - 4x + 4 = (x-2)^2.$

Therefore, we have that $x-2 = \pm 16^{3/2} = \pm 64$.  Therefore, the possible values of $x$ are $-62$ and $66,$ and the only positive value is therefore $\boxed{66}$.
<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Solve for the positive value of $x$ such that $\sqrt[3]{x^2 - 4x + 4} = 16$.
Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
<|im_start|>assistant

user
66
Solve for the positive value of $x$ such that $\sqrt[3]{x^2 - 4x + 4} = 16$.
test