The domain of definition of the function $f(x)=\sqrt[6]{4^x+8^{\frac{2}{3}(x-2)}-52-2^{2(x-1)}}$ is:

$[3, \infty)$

$4^x+8^{\frac{2}{3}(x-2)}-52-2^{2(x-1)}>0$

$⇒2^{2x}+2^{2(x-2)}-52-2^{2(x-1)}>0$

$⇒2^{2x}\left[1+\frac{1}{16}-\frac{1}{4}\right]>52⇒2^{2x}\left[\frac{16+1-4}{16}\right]>52$

so $2^{2x}\left[\frac{13}{16}\right]>52$

$2^{2x}>64$

so $x ≥3 ⇒x∈[3, \infty)$

Hence (3) is the correct answer.