Find the values for $a,b,c,d$ 
Given  $$x^3+4x^2y+axy^2+3xy-bx^cy+7xy^2+dxy+y^2=x^3+y^2$$  for any
  real number $x$ and $y$ , find the value of $a,b,c,d$


$$x^3+4x^2y+axy^2+3xy-bx^cy+7xy^2+dxy+y^2=x^3+y^2$$
cancel out $x^3$ and $y^2$
our equation becomes
$$4x^2y+axy^2+3xy-bx^cy+7xy^2+dxy=0$$
write it as following
$$4x^2y-bx^cy+axy^2+7xy^2+dxy+3xy=0$$
take common terms 
$$y(4x^2-bx^c)+xy^2(a+7)+xy(d+3)=0$$
this equation has to be true for any real x and y thus 
you can clearly see these terms has to cancel out each other 
$$4x^2-bx^c=0$$
$$a+7=0$$
$$d+3=0$$
we obtain 

$$a=-7$$
  $$ b=4$$ $$c=2$$  $$d=-3$$

