Here is the statement :
a) Find the smallest number that can be written as the sum of 2 cubes in 2 different ways.
$\textit{Tips : Use a generative function}$
Ok so at the beginning I wrote : $\left\{\begin{matrix} n=\alpha^{3} + \beta ^{3}
\\ n=\gamma^{3} + \epsilon ^{3}
\end{matrix}\right.$
And then to consider generative function I said that $\left\{\begin{matrix} x^{n}=x^{\alpha^{3}+\beta ^{3}}
\\ x^{n}=x^{\gamma^{3}+\epsilon ^{3}}
\end{matrix}\right.$
So now I can play with the multiplication and so : $\begin{align*} x^{n}=x^{\alpha^{3}+\beta ^{3}}=x^{\gamma^{3}+\epsilon ^{3}}\\ x^{n}=x^{\alpha^{3}}x^{\beta^{3}}=x^{\gamma^{3}}x^{\epsilon ^{3}} \end{align*}$
So to find $n$, I need to do the generative function of $\alpha ,\beta ,\gamma ,\epsilon $.
So I wrote the generative function of $\alpha$ as $\sum_{0}^{n}x^{\alpha^{3}}$ and for the others too.
Then my idea was to multiply them like : $\sum_{0}^{n}x^{\beta^{3}}*\sum_{0}^{n}x^{\alpha^{3}}=\sum_{0}^{n}x^{\gamma^{3}}*\sum_{0}^{n}x^{\epsilon^{3}}$
And then where I found the same exponent in both of this new polynomial, it would be $n$. But of course it's not because the $2$ new polynomial that I made are exactly equal.
I need to find a range for $\alpha ,\beta ,\gamma ,\epsilon $ so when I found the same power of x in both of new polynomial, I know it is the result of $\alpha^{3} + \beta^{3} = \gamma^{3} + \epsilon^{3}$ with $(\alpha,\beta) \neq (\gamma,\epsilon)$