How to show that $\int_0^1 \left(\sqrt[3]{1-x^7} - \sqrt[7]{1-x^3}\right)\;dx = 0$ 
Evaluate the integral: $$ \int_0^1 \left(\sqrt[3]{1-x^7} - \sqrt[7]{1-x^3}\right)\;dx$$

The answer is $0,$ but I am unable to get it. There is some symmetry I can not see.

Observing the inverse of the function $f(x) = \sqrt[3]{1 - x^7}$ on the interval $[0,1]$ is $f^{-1} (x) = \sqrt[7]{1 - x^3}$, using the result
$$\int^b_a f(x) \, dx + \int^{f(b)}_{f(a)} f^{-1} (x) \, dx = b f(b) - a f(a),$$
since $f(a) = f(0) = 1$ and $f(b) = f(1) = 0$ it immediately follows that
$$\int^1_0 \sqrt[3]{1 - x^7} \, dx + \int^0_1 \sqrt[7]{1 - x^3} \, dx = 0,$$
or
$$\int^1_0 \left \{\sqrt[3]{1 - x^7} - \sqrt[7]{1 - x^3} \right \}\, dx = 0.$$
