Time to be serious though...(Nod)
I would first arrange the equation as:
$$(7x+1)^{\frac{1}{3}}+\left(x^2-8x-1 \right)^{\frac{1}{3}}=2+\left(x^2-x-8 \right)^{\frac{1}{3}}$$
Let:
$$u=7x+1$$
$$v=x^2-8x-1$$
$$w=x^2-x-8$$
and we have:
(1) $$u^{\frac{1}{3}}+v^{\frac{1}{3}}=2+w^{\frac{1}{3}}$$
Cubing both sides of the equation, we find after simplification:
$$(uv)^{\frac{1}{3}}\left(u^{\frac{1}{3}}+v^{\frac{1}{3}} \right)=2w^{\frac{1}{3}}\left(2+w^{\frac{1}{3}} \right)$$
Using (1), we may write:
$$(uv)^{\frac{1}{3}}\left(u^{\frac{1}{3}}+v^{\frac{1}{3}} \right)=2w^{\frac{1}{3}}\left(u^{\frac{1}{3}}+v^{ \frac{1}{3}} \right)$$
We may arrange this as:
$$\left(u^{\frac{1}{3}}+v^{\frac{1}{3}} \right)\left((uv)^{\frac{1}{3}}-2w^{\frac{1}{3}} \right)=0$$
Using the zero-factor property, this yields the cases:
i) $$u^{\frac{1}{3}}+v^{\frac{1}{3}}=0$$
$$u=-v$$
Back-substituting for $u$ and $v$, we find:
$$7x+1=-x^2+8x+1$$
$$x^2-x=0$$
$$x(x-1)=0$$
$$x=0,1$$
ii) $$(uv)^{\frac{1}{3}}-2w^{\frac{1}{3}}=0$$
$$uv=8w$$
Back substituting for $u$, $v$, and $w$, we find:
$$(7x+1)\left(x^2-8x-1 \right)=8\left(x^2-x-8 \right)$$
After simplification, we obtain:
$$(x+1)(x-1)(x-9)=0$$
$$x=-1,1,9$$
we have already ascertained that the four solutions we found are valid, hence we may state the real solutions are:
$$x=-1,0,1,9$$