$\sqrt[3]{10-x}+\sqrt[3]{30-x}=\sqrt[3]{15-x}+\sqrt[3]{25-x}$

Question

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.

Answer 1

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 $x=20$

Answer 2

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 !