show that $\int_0^1 (\sqrt[3]{1-x^7}-\sqrt[7]{1-x^3}) \, dx=0$ show that
$$I=\int_0^1 (\sqrt[3]{1-x^7}-\sqrt[7]{1-x^3}) \, dx=0$$
I find this is Nice equalition!
My try: let
$$\sqrt[3]{1-x^7}=t\Longrightarrow x=\sqrt[7]{1-t^3}$$
so
$$dx=-\dfrac{3}{7}t^2(1-t^3)^{-\dfrac{6}{7}} \, dt$$
so
$$I=\frac{3}{7}\int_0^1 \frac{t^3}{\sqrt[7]{(1-t^3)^6}} \, dt-\int_0^1 \sqrt[7]{1-x^3} \, dx$$
By parts,we have
$$\int_0^1 \sqrt[7]{1-x^3} dx=\dfrac{3}{7}\int_0^1 \frac{x^3}{\sqrt[7]{(1-x^3)^6}} \, dt$$
so
$$I=0$$
this problem maybe have more other nice methods!Thank you

You can write the two parts of the integrals as follows:
$$
\int_0^1 \sqrt[3]{1-x^7} \, dx = \int_0^1 \int_0^\sqrt[3]{1-x^7} \, dy\,dx
$$
and
$$
\int_0^1 \sqrt[7]{1-y^3}) \, dy = \int_0^1 \int_0^\sqrt[7]{1-y^3} \, dx\,dy.
$$
Now you can show that the regions of both these integral are the same. Namely the sets $\{(x, y): 0 \leq x \leq 1, 0 \leq \sqrt[3]{1-x^7}\}$ and $\{(x, y): 0 \leq y \leq 1, 0 \leq x \leq \sqrt[7]{1-y^3}\}$ are the same.
