An algebra problem by arko roychoudhury

$\sqrt{\frac{8^{10} + 4^{10}}{8^4+4^{11}}} = \sqrt{\frac{2^{30} + 2^{20}}{2^{12} + 2^{22}}} = \sqrt{\frac{(2^{20})(2^{10}+1)}{(2^{12})(2^{10} + 1)}}$

$= \frac{2^{20/2}}{2^{12/2}} =\frac{2^{10}}{2^6} = 2^{10-6} = 2^4 =\boxed{16}$ .

Well that took just 30 seconds

Common factor in numerator under the square-root is 4^10 and in the denominator it is 8^4, that being 2^20 and 2^12, respectively which equals 2^8. When brought out it is 2^4 which is 16. The interesting thing is that the remaining factors within square-root in numerator and denominator being equal cancel out.