Find the values for $a,b,c,d$

Question

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 $a,b,c,d$

Answer 1

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

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 $b=4.$

Answer 2

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

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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 $d=-3$.

Answer 3

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 $\color{blue}{b=4}$.

Answer 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

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