Let $N=a^3+b^3$. We take $y=t+b$, $x= a-\frac{b^2}{a^2}\cdot {t}$, assume $y^3+x^3=N$ and find $t=\frac {3a^{3}b}{b^3-a^3}$. It gives $y=b+\frac{3a^3b}{b^3-a^3}$, $x=a-\frac{3ab^3}{b^3-a^3}$. For example, $1^3+2^3=9$, following positive solution $(\frac{1243617733990094836481}{609623835676137297449})^3+(\frac{487267171714352336560}{609623835676137297449})^3=9$.

juna wrote: Let $N=a^3+b^3$. We take $y=t+b$, $x= a-\frac{b^2}{a^2}\cdot {t}$, What is the motivation behind doing this?

SoumikNayak wrote: juna wrote: Let $N=a^3+b^3$. We take $y=t+b$, $x= a-\frac{b^2}{a^2}\cdot {t}$, What is the motivation behind doing this? I can't find the exact source but there is some discussion here. https://artofproblemsolving.com/community/q2h54186p9111194

Nice problem. Let $(a,b)$ be a rational solution to $x^3+y^3=9$. Consider the curve $x^3+y^3=9$ and the point $(a,b)$ on it. The tangent from $(a,b)$ to the curve is the line $a^2x+b^2y=9$. Solving for the coordinates the other intersection of this tangent with the curve leads to the third degree equation $(x-a)^2((b^3-a^3)x+a(9+b^3))=0$ for the $x$ coordinate. This gives another rational point $(\frac{-a(9+b^3)}{b^3-a^3}, \frac{b(9+a^3)}{b^3-a^3})$. Starting from the obvious solution $(1,2)$, we get infinitely many solutions this way.