$\sqrt[3]{10-x}+\sqrt[3]{30-x}=\sqrt[3]{15-x}+\sqrt[3]{25-x}$ I just happened to find a problem and an elegant solution.
The question asks us to solve the following equation
$$\sqrt[3]{10-x}+\sqrt[3]{30-x}=\sqrt[3]{15-x}+\sqrt[3]{25-x}$$
I am answering this question below but I would love if you can also share a different solution.
P.S: I composed this problem by myself. I do not know if this problem is available anywhere. I would love to get some feedback about the same. It motivates me to create problems and discuss with others.

Suppose $$A=\sqrt[3]{10-x}+\sqrt[3]{30-x}=\sqrt[3]{15-x}+\sqrt[3]{25-x}$$
I will use the following
\begin{align}
A^3=&(p+q)^3 \\
 =&p^3+q^3+3pq(p+q) \\
=&p^3+q^3+3pq(A)
\end{align}
Then
\begin{align}
A^3=&(10-x)+(30-x)+3(\sqrt[3]{10-x})(\sqrt[3]{30-x})(A) \\
=&(40-2x)+3(\sqrt[3]{10-x})(\sqrt[3]{30-x})(A) 
\end{align}
\begin{align}
A^3=&(15-x)+(25-x)+3(\sqrt[3]{15-x})(\sqrt[3]{25-x})(A) \\
=&(40-2x)+3(\sqrt[3]{15-x})(\sqrt[3]{25-x})(A) 
\end{align}
and hence we have
$$(\sqrt[3]{10-x})(\sqrt[3]{30-x})(A)=(\sqrt[3]{15-x})(\sqrt[3]{25-x})(A)$$
which is impossible unless $A=0$
Hence $$A=\sqrt[3]{10-x}+\sqrt[3]{30-x}=0$$ and
I conclude $x=20$
