# Are the binomial coefficients unique?

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?

• Is that equation supposed to be true for all $a,b,x,y$? – YiFan Apr 14 at 22:26
• Not necessarily. – NumThcurious Apr 14 at 23:39
• then obviously the answer is yes. Choose $a=b=x=y=0$ and any $u,v,w,t$ works. – YiFan Apr 15 at 1:12
• I did mention that $$ax+by\neq 0$$ – NumThcurious Apr 15 at 15:15

$$ux^3+vx^2y+wxy^2+ty^3=a^3x^3+3a^2byx^2+3ab^2y^2x+b^3y^3$$ is an equaliy involving two polynomials respect to x. since $$( x^3, x^2, x, 1)$$ is a basis then, $$(u,v,w,t)= ( a^3, 3a^2b, 3ab^2, b^3)$$