$\sqrt[3]{10-x}+\sqrt[3]{30-x}=\sqrt[3]{15-x}+\sqrt[3]{25-x}$ 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.
 A: 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 $x=20$
A: Let $\sqrt[3]{10-x}=a,\sqrt[3]{30-x}=b,\sqrt[3]{15-x}=c, \sqrt[3]{25-x}=d$
We have $a+b=c+d\ \ \ \ (1)$
Again $a^3+b^3=c^3+d^3$
$\iff(a+b)^3-3ab(a+b)=(c+d)^3-3cd(c+d)$
So, either case$\#1: a+b=0$
or case$\#2:ab=cd$
in that case, $(a,b); (c,d)$ are the roots of the same quadratic equation
$\implies$ either $a=c, b=d$ or $a=d,b=c$
Can you take it from here?
Btw, thanks for posting the nice problem !
