Find the remainder $1690^{2608} + 2608^{1690}$ when divided by 7? Find the remainder $1690^{2608} + 2608^{1690}$ when divided by 7?

My approach:-
$1690 \equiv 3(\bmod 7)$
$1690^{2} \equiv 2(\bmod 7)$
$1690^{3} \equiv-1 \quad(\mathrm{mod} 7)$[ quite easy to determine , $\frac{2*1690}{7}$..so on]
$\left(1690^{3}\right)^{869} \cdot 1690 \equiv(-1)^{869}1690 \quad(\mathrm{mod} 7)$
$1690^{2608} \equiv -1690 \quad(\mathrm{mod} 7)$....(1)
again for $2608$
$2608 \equiv 4(\bmod 7)$
$2608^{2} \equiv 2(\bmod 7)$
$2608^{3} \equiv1 \quad(\mathrm{mod} 7)$[ quite easy to determine , $\frac{2*2608}{7}$..so on]
$\left(2608^{3}\right)^{563} \cdot 2608 \equiv(1)^{563}2608 \quad(\mathrm{mod} 7)$
$2608^{1690} \equiv 2608 \quad(\mathrm{mod} 7)$...(2)
Now applying property
adding (1) + (2),
$1690^{2608} + 2608^{1690}=918 \quad(\mathrm{mod} 7)$
$\boxed{1690^{2608} + 2608^{1690} \equiv 1 \quad(\mathrm{mod} 7)}$
Is my approach best? or Anyother approach is there comparatively better than it

A shorter way would be to use Fermat's little theorem, plus the fact that $1690 = 7k+3$ and $2608 = 7k-3$
$$1690^6 \equiv 1 \text{   (mod 7)} \implies 1690^{2608} \equiv 1690^4 \text{ (mod 7)} \equiv 3^4 \text{(mod 7)} \equiv 4 \text{(mod 7)}$$
Similarily,
$$2608^6 \equiv 1 \text{   (mod 7)} \implies 2608^{1690} \equiv 2608^4 \text{ (mod 7)} \equiv 3^4 \text{(mod 7)} \equiv 4 \text{ (mod 7)}$$
Adding the remainders, $4 + 4 \text{  (mod 7)} \equiv 1 \text{  (mod 7)}$
