Last three digits of $\sqrt{4^{2016}+2\cdot 6^{2016}+9^{2016}}$ Find the last three digits of $$\sqrt{4^{2016}+2\cdot 6^{2016}+9^{2016}}$$
I don't know how to continue my work:
$$\sqrt{4^{2016}+2\cdot 6^{2016}+9^{2016}}=\sqrt{2^{2(2016)}+2\cdot (3\cdot 2)^{2016}+3^{2(2016)}}=\sqrt{2^{4032}+2\cdot 3^{2016}\cdot 2^{2016}+3^{4032}}=\sqrt{2^{4032}+2^{2017}\cdot 3^{2016}+3^{4032}}=?$$

If you follow the hint from @user35508, lets make a congruence.
$$2^{2016}+3^{2016} \text{ mód } 1000$$
Thanks to the Euler's totient function, we know that $a^{\phi(b)}\equiv 1 \text{ mod } b$ if mcd(a,b)=1. The factorization of $1000 = (5*2)^3$ so its trivial that $mcd(2,3,1000)=1$.
$$\phi(1000)=\phi((5*2)^3)=\phi(5^3)\cdot\phi(2^3)=(5^3-5^2)\cdot(2^3-2^2)=400$$
Finally, lets divide $2016/400 = 5\cdot 400 + 16$ and apply it to the first formula.
$$2^{5*400+16}+3^{5*400+16}\text{ mod } 1000 = (2^5)^{400}\cdot{2^{16}}+(3^5)^{400}\cdot{3^{16}}\text{ mod } 1000 = 2^{16}+3^{16}\text{ mod } 1000$$
The remainder is $43112257\text{ mod } 1000 = 257$ that is also the three last digits.
