Let $P(x) = x^3+(x+1)^3+(x+2)^3+\cdots+(x+7)^3$. Expanding the right side yields $$P(x) = 8x^3+84x^2+420x+784$$

• If $x\ge 0$ then $(2x+7)^3 < P(x) < (2x+10)^3$. This means that $y^3$ is between $(2x+7)^3$ and $(2x+10)^3$. Hence, $y=2x+8$ or $y=2x+9$.
  • $y=2x+8 \implies 8x^3+84x^2+420x+784 = (2x+8)^3$
  • $y=2x+9 \implies 8x^3 + 84x^2 + 420x +784 = (2x+9)^3$. Both equations do not have integer solutions.
• If $x < 0$, we note that if $(x, y)$ is a solution, so will be $(-x-7, -y)$. Therefore there is no solution when $x\le -7$. Try $x=-1, -2, -3, \cdots, -6$, respectively, find the following solutions: $$\boxed{(-2,6), (-3, 4), (-4, -4), (-5, -6)}$$