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Let $a,x,b,y$ be integers. Can we find rationals $u,v,w,t$ such that $$(ax+by)^3=ux^3+vx^2y+wxy^2+ty^3\neq 0$$ where $$(u,v,w,t)\neq ( 1, 3a^2b, 3ab^2, 1)$$ The answer looks trivial but can one prove it?

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  • $\begingroup$ Is that equation supposed to be true for all $a,b,x,y$? $\endgroup$ – YiFan Apr 14 at 22:26
  • $\begingroup$ Not necessarily. $\endgroup$ – NumThcurious Apr 14 at 23:39
  • $\begingroup$ then obviously the answer is yes. Choose $a=b=x=y=0$ and any $u,v,w,t$ works. $\endgroup$ – YiFan Apr 15 at 1:12
  • $\begingroup$ I did mention that $$ax+by\neq 0$$ $\endgroup$ – NumThcurious Apr 15 at 15:15
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$$ux^3+vx^2y+wxy^2+ty^3=a^3x^3+3a^2byx^2+3ab^2y^2x+b^3y^3$$ is an equaliy involving two polynomials respect to x. since $ ( x^3, x^2, x, 1)$ is a basis then, $$(u,v,w,t)= ( a^3, 3a^2b, 3ab^2, b^3)$$

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