1. In the set of real numbers, solve the equation:

$$
(x-7)^{3}+(x+3)^{3}=278(x-2)
$$
1. Based on the identity $a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)$, we can factorize the left side, and thus we are left with the equation

$$
(2 x-4)\left((x-7)^{2}-(x-7)(x+3)+(x+3)^{2}\right)=278(x-2) .
$$

One solution is obviously $x_{1}=2$. We will seek the other solutions after canceling both sides by $2(x-2)$: on the left side, we then have $\left(x^{2}-14 x+\right.$ 49) $-\left(x^{2}-4 x-21\right)+\left(x^{2}+6 x+9\right)=x^{2}-4 x+79$, and on the right side, we have 139. Therefore, we need to solve the quadratic equation $x^{2}-4 x-60=0$, and its solutions are $x_{2 / 3}=\frac{4 \pm \sqrt{16+240}}{2}=\frac{4 \pm 16}{2}=2 \pm 8$, i.e., $x_{2}=-6$ and $x_{3}=10$.