3. The original equation is equivalent to $\left(x^{2}+x+1\right)^{2} \equiv 0\left(\bmod 7^{6}\right)$, and further equivalent to $x^{2}+x+1 \equiv$ $0\left(\bmod 7^{3}\right), x \equiv-19,18\left(\bmod 7^{3}\right)$. (These are all the solutions, not the number of solutions).
3. Solve the congruence equation
$$(x+1)^{7}-x^{7}-1 \equiv 0\left(\bmod 7^{7}\right)$$

for solutions satisfying $7 \nmid x(1+x)$.