Vtyjkyuk

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 $a,b,c,d$

$$x^3+4x^2y+axy^2+3xy-bx^cy+7xy^2+dxy+y^2=x^3+y^2$$

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 $b=4.$

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 $yne 0$ we get $$4x^2-bx^c+(d+3)x=0;;;forall x$$

a) Put $x= 2$: $$boxed8+b(2-2^c)=0$$

b) Put $x=4$: $$boxed48+b(4-4^c)=0$$

Solving this system we get $$-b = 8over 2-2^c = 48over 4-4^cimplies 2+2^c= 6implies c = 2$$ so $b=4$ and $d=-3$.

Starting from the equation
beginalign*
x^3+4x^2y+axy^2+3xy-bx^cy+7xy^2+dxy+y^2=x^3+y^2
endalign*
we could go on as follows.

First step: Simplify and put each terms to the left-hand side. We obtain
beginalign*
4x^2y+axy^2+3xy-bx^cy+7xy^2+dxy=0tag1
endalign*
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
beginalign*
(4x^2+3x-bx^c+dx)colorbluey+(ax+7x)colorbluey^2=0
endalign*

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:
beginalign*
ax+7x=0
endalign*
Since the coefficient of $x$ has to be zero we see $colorbluea=-7$.

The other expression is
beginalign*
4x^2+3x-bx^c+dx=0
endalign*
We have a linear term in $x$, namely $3x+dx=0$ and conclude $colorblued=-3$. We see that setting $colorbluec=2$ gives a second quadratic term $4x^2-bx^2=0$ and we finally conclude $colorblueb=4$.

$$x^3+4x^2y+axy^2+3xy-bx^cy+7xy^2+dxy+y^2=x^3+y^2$$

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

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