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.
 A: Let $m, n > 0$. Then observe that
$$ \int_{0}^{1} \sqrt[n]{1-x^m} \; dx$$
is the area of the region given by inequalities
$$ 0 \leq x \leq 1 \quad \text{and} \quad 0 \leq y \leq \sqrt[n]{1-x^m}.$$
But the last inequality is equivalent to $0 \leq x^m + y^n \leq 1$. Thus
$$ \int_{0}^{1} \sqrt[n]{1-x^m} \; dx = [\text{Area given by} \ 0 \leq x^m + y^n \leq 1, \ 0 \leq x, y \leq 1 ]$$
Thus by interchanging the role of $x$ and $y$, we have
$$ \int_{0}^{1} \sqrt[n]{1-x^m} \; dx = \int_{0}^{1} \sqrt[m]{1-x^n} \; dx.$$

Of course, we can give a purely analytic approach. Let $y = \sqrt[3]{1 - x^7}$. Then $x = \sqrt[7]{1 - y^3}$ and hence by integration by substitution,
$$\begin{align*}
\int_{0}^{1} \sqrt[3]{1 - x^7} \; dx
&= \int_{0}^{1} y(x) \; dx \\
&= \int_{1}^{0} y \; dx(y) \\
&= [y x(y)]_{1}^{0} - \int_{1}^{0} x(y) \; dy \\
&= \int_{0}^{1} \sqrt[7]{1 - y^3} \; dy.
\end{align*}$$
A: Another way to go is to use $\beta$-function

$$ \mathrm{\beta}(x,y) = \int_0^1t^{x-1}(1-t)^{y-1}\,dt,\quad {Re}(x),{Re(y)}>0.  $$

$$ \int_0^1 \sqrt[3]{1-x^7} \mathrm dx = \frac{1}{7}\int_0^1 u^{-6/7}(1-u)^{1/3} \mathrm dx=\dots\,. $$
Now, you can finish the problem.
A: 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
$$\int^1_0 \left \{\sqrt[3]{1 - x^7} - \sqrt[7]{1 - x^3} \right \}\, dx = 0.$$
