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 $a,b,c,d$

 A: Put some values of $x$ and $y$ in it.
Say $x=1$:  $$(a+7)y^2+(7-b+d)y =0\;\;\forall y$$ so $a=-7$ and $b-d=7$. 
Now we have (for all $x,y$):
$$4x^2y-bx^cy+(d+3)xy=0$$
If $y\ne 0$ we get $$4x^2-bx^c+(d+3)x=0\;\;\;\forall x$$

a) Put $x= 2$: $$\boxed{8+b(2-2^c)=0}$$
b) Put $x=4$: $$\boxed{48+b(4-4^c)=0}$$
Solving this system we get $$-b = {8\over 2-2^c} = {48\over 4-4^c}\implies 2+2^c= 6\implies c = 2$$ so $b=4$ and $d=-3$.
A: Starting from the equation
\begin{align*}
x^3+4x^2y+axy^2+3xy-bx^cy+7xy^2+dxy+y^2=x^3+y^2
\end{align*}
we could go on as follows.

First step: Simplify and put each terms to the left-hand side. We obtain
  \begin{align*}
4x^2y+axy^2+3xy-bx^cy+7xy^2+dxy=0\tag{1}
\end{align*}
  by noting that the terms $x^3$ and $y^2$ cancel out.

We have in (1) a bivariate polynomial in $x$ and $y$ at the left-hand side and the zero polynomial at the right-hand side. We have equality with the zero-polynomial iff the coefficients of $x^ky^l$ are zero for each $k,l$ at the left-hand side.
The next step is to either arrange the left-hand side by increasing powers of $x$ or by increasing powers of $y$. Since there is a somewhat tricky unknown $c$, which is a power of $x$ and which makes arrangement by increasing powers of $x$ cumbersome, we decide to collect according to increasing powers of $y$.

Second step: By arranging according to increasing powers of $y$ we obtain
  \begin{align*}
(4x^2+3x-bx^c+dx)\color{blue}{y}+(ax+7x)\color{blue}{y^2}=0
\end{align*} 

Now we are ready to harvest. We can now easily determine the unknowns $a,b,c,d$ by comparing corresponding powers of $x$.

Third step: Compare corresponding powers of $x$.
We start with the easy one:
  \begin{align*}
ax+7x=0
\end{align*}
  Since the coefficient of $x$ has to be zero we see $\color{blue}{a=-7}$.
The other expression is 
  \begin{align*}
4x^2+3x-bx^c+dx=0
\end{align*}
  We have a linear term in $x$, namely $3x+dx=0$ and conclude $\color{blue}{d=-3}$. We see that setting $\color{blue}{c=2}$ gives a second quadratic term $4x^2-bx^2=0$ and we finally conclude $\color{blue}{b=4}$.

A: $$x^3+4x^2y+axy^2+3xy-bx^cy+7xy^2+dxy+y^2=x^3+y^2$$
R.H.S only has $x^3$ and $y^2$, therefore all the terms in L.H.S must cancel out except for $x^3$ and $y^2$.
So let's pair the terms that may cancel out.
$$x^3+y^2+(axy^2+7xy^2)+(3xy+dxy)+4x^2y-bx^cy$$
From here you can conclude that,
$a=-7,d=-3$
Also for $4x^2y$ to cancel $c=2$ and $b=4.$
A: $$x^3+4x^2y+axy^2+3xy-bx^cy+7xy^2+dxy+y^2=x^3+y^2$$
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 

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

