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

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 $$ \int_{0}^{1} \sqrt[n]{1-x^m} \; dx = \int_{0}^{1} \sqrt[m]{1-x^n} \; dx.$$

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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, $$\begin{align*} \int_{0}^{1} \sqrt[3]{1 - x^7} \; dx &= \int_{0}^{1} y(x) \; dx \\ &= \int_{1}^{0} y \; dx(y) \\ &= [y x(y)]_{1}^{0} - \int_{1}^{0} x(y) \; dy \\ &= \int_{0}^{1} \sqrt[7]{1 - y^3} \; dy. \end{align*}$$

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.

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 $$\int^1_0 \left \{\sqrt[3]{1 - x^7} - \sqrt[7]{1 - x^3} \right \}\, dx = 0.$$