$Q.$ Solve for $x$ , If $(a,x)\in\mathbb N $ s.t. $$(a+x)^{\frac{1}{3}}+(a-x)^{\frac{1}{3}}=3(a^2-x^2)^{\frac{1}{3}}$$
$\Rightarrow$ My approach was to cube both the sides : First let , $\underbrace{3(a^2-x^2)^{\frac{1}{3}}}_\Lambda$ So , by cubing both sides we get , $$\Lambda^3-\Lambda^2=2a$$
If $(a^2-x^2)$ is a whole cube then $3(a^2-x^2)^{\frac{1}{3}}\in\mathbb N$ now according to the above condition , i.e. $\Lambda^3-\Lambda^2=2a$
hence , $a\in\{2,9,24,50,.......\frac{n^2(n-1)}{2}\}$
I'm not able to proceed form here , Much thank you
I claim that there are no solutions $x,a$ in positive integers.
To prove this, write $a+x=m$ and $a-x=n$ with $m\in \Bbb N$ and $n\in \Bbb Z$. Then the equation is $$ m^{1/3}+n^{1/3}=3(mn)^{1/3}. $$
We have $n<m$ because of $x>0$. Also $n=0$ gives $a=x$, which is a contradiction to the equation. The equation is equivalent to $$ 1+\left(\frac{n}{m}\right)^{1/3}=3n^{1/3}. $$ Suppose that $n$ is positive. Then the LHS is less than $2$, since $n<m$. So we obtain $$ n^{1/3}<\frac{2}{3}, $$ which gives $n<1$. Since $n$ is positive, we have $n=0$, a contradiction. So $n$ is negative and the RHS of our original equation is negative, but the LHS is positive, because of $$ (a+x)^{1/3}+(a-x)^{1/3}>0 $$ for $a,x$ positive integers.