Let $A=k[x,y,z]$, and let $I$ be generated by $(Ax+Ay)(Ax+Az)(Ay+Az)$. I wish to find a set of three generators for $I$.
My first approach for this was by expanding out. It seems that by expanding this, I get that $I=Ax^2+Ay^2+Az^2+Axy+Ayz+Axz$, implying that if $f(x,y,z)=x+y+z$, then $f^2\in I$. But I know that $\forall\ f\in I,\ \forall\ x,y,z\in k, f(x,0,0)=f(0,y,0)=f(0,0,z)=0$, but this is clearly a contradiction.
Where did this approach go wrong?
Try your expansion again. Each term will be A times a monomial of degree 3. This will fix things.