278. Solve the equation:

$$
(x+3)^{4}+(x+5)^{4}=16
$$
$\triangle$ Let $x+4=y$. Why exactly $x+4$? Here's where it comes from:

$$
\frac{(x+3)+(x+5)}{2}=x+4
$$

We have:

$$
(y-1)^{4}+(y+1)^{4}=16
$$

Now we need to square $y-1$ and $y+1$ in the left part of the equation, and then square the result again. After simplifications, a biquadratic equation is formed:

$$
y^{4}+6 y^{2}-7=0
$$

Its roots are $y_{1,2}= \pm 1$. From here, it is easy to find $x$.

Answer: $-3,-5$.