A giant root

Notice that $2013 \cdot 2=4026$ . Therefore, we first find the square root of $49-20\sqrt{6}$ . Let this value be $a-b\sqrt{c}$ . Squaring both, we get $49-20\sqrt{6}=a^2+b^2c-2ab\sqrt{c}$ . This shows that $49=a^2+b^2c$ and $20\sqrt{6}=2ab\sqrt{c}$ . This obviously shows that $c=6$ . The first equation now reads $a^2+6b^2=49$ , while simplifying the second one gives $ab=10$ . We can easily test different values and find that $a=5$ and $b=2$ . Therefore, $\sqrt{49-20\sqrt{6}}=\sqrt{a-b\sqrt{c}}=\sqrt{5-2\sqrt{6}}$ . Our original equation turns to $\sqrt[2013]{5+2\sqrt{6}} \times \sqrt[2013]{5-2\sqrt{6}}$ . Multiplying them, we get $\sqrt[2013]{(5+2\sqrt{6})(5-2\sqrt{6})}=\sqrt[2013]{25-24}=\sqrt[2013]{1}=\sqrt{\boxed{1}}$ .