<s>[INST] 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
 [/INST] The antiderivative of $(1-x^a)^{1/b}$, at least for integer values of $a$ and $b$, is given by $x{\rm Hypergeometric2F1}[1/a, -(1/b), 1 + 1/a, x^a]$.
The integral of $(1-x^a)^{1/b}$ between 0 and 1 is then given by
$(\Gamma[1 + 1/a] \Gamma[1 + 1/b]) / \Gamma[1 + 1/a + 1/b]$
Then, the two integrals are equal and your identity is proved.     
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