Find all integer solutions of equality

Question

Find all integer solutions of equation $$x^3+(x+1)^3+...+(x+7)^3=y^3$$ I've solved it by opening brackets and consideration of signs but I think there is simpler way of solving it .

Answer 1

Let $$P(x)=x^3+(x+1)^3+(x+2)^3+(x+3)^3+(x+4)^3+(x+5)^3+(x+6)^3+(x+7)^3$$ so $$P(-x-7)=-P(x)$$

on the other hand we can $$P(x)=8x^3+84x^2+420x+784$$ if $x\ge 0$,we have $$(2x+7)^3=8x^3+84x^2+294x+343<P(x)<8x^3+120x^2+600x+1000=(2x+10)^3$$ so we have $$(2x+7)<y<2x+10$$ so $$P(x)=(2x+8)^3\Longrightarrow -12x^2+36x+272=0$$ this equation no integer roots,

simaler we have $$P(x)-(2x+9)^3=-24x^2-66x+55=0$$ also have no integer roots.

Now we have prove $x\ge 0 $ or $x\le -7$ this equation have no integer roots

But if $-6\le x\le 1$, we can easy to find when $x=-2\Longrightarrow P(-2)=216=6^3$

$x=-3\Longrightarrow P(-3)=64=4^3$

$x=-4,P(-4)=-64=(-4)^3$

$x=-5,P(-5)=216^3=(-6)^3$