Partial fraction decomposition

[Drain Brain]

please help decompose$\frac{4x^2y}{(x^2-2xy+2y^2)(x^2+2xy+2y^2)}$

I've used the cases I know for this problem but to no avail. please help me.

 

[mathbalarka]

This takes a bit of trickery, note that :

$$\begin{align}\frac{4x^2y}{(x^2-2xy+2y^2)(x^2+2xy+2y^2)} &= \frac{x \cdot 4xy}{(x^2-2xy+2y^2)(x^2+2xy+2y^2)} \\ &= x \cdot \frac{(x^2+2xy+2y^2) - (x^2-2xy+2y^2)}{(x^2-2xy+2y^2)(x^2+2xy+2y^2)}\end{align}$$

Can you proceed?

 

[Drain Brain]

This takes a bit of trickery, note that :

$$\begin{align}\frac{4x^2y}{(x^2-2xy+2y^2)(x^2+2xy+2y^2)} &= \frac{x \cdot 4xy}{(x^2-2xy+2y^2)(x^2+2xy+2y^2)} \\ &= x \cdot \frac{(x^2+2xy+2y^2) - (x^2-2xy+2y^2)}{(x^2-2xy+2y^2)(x^2+2xy+2y^2)}\end{align}$$

Can you proceed?

sure, from here It seems that I can decompose it. but how come you replaced $4xy$ to $(x^2+2xy+2y^2) - (x^2-2xy+2y^2)$ ??

 

[mathbalarka]

Uh, not sure if I understand your question, but it follows from basic algebra

$$(x^2+2xy+2y^2) - (x^2-2xy+2y^2) = \cancel{\color{red}{x^2}} + 2xy + \cancel{\color{green}{2y^2}} - \cancel{\color{red}{x^2}} + 2xy - \cancel{\color{green}{2y^2}} = 2xy + 2xy = \boxed{4xy}$$

 

[MarkFL]

While mathbalarka's suggestion is quite clever and makes light work of the problem, I would have assumed the decomposition would take the form:

\(\displaystyle \frac{4x^2y}{\left(x^2-2xy+2y^2\right)\left(x^2+2xy+2y^2\right)}=\frac{Ax+By+C}{x^2-2xy+2y^2}+\frac{Dx+Ey+F}{x^2+2xy+2y^2}\)

and then plodded along with the resulting cumbersome algebra.

Hence:

\(\displaystyle 4x^2y=(Ax+By+C)\left(x^2+2xy+2y^2\right)+(Dx+Ey+F)\left(x^2-2xy+2y^2\right)\)

\(\displaystyle 4x^2y=(A+D)x^3+(C+F)x^2+(2A+B-2D+E)x^2y+(2A+2B+2D-2E)xy^2+(2C-2F)xy+(2C+2F)y^2+(2B+2E)y^3\)

Comparing coefficients, we obtain:

\(\displaystyle A+D=0\)

\(\displaystyle C+F=0\)

\(\displaystyle 2A+B-2D+E=4\)

\(\displaystyle A+B+D-E=0\)

\(\displaystyle C-F=0\)

\(\displaystyle B+E=0\)

From the 2nd and 5th equations, we immediately find:

\(\displaystyle C=F=0\)

From the 1st, 4th and 6th, we find:

\(\displaystyle B=E=0\)

Thus, we are left with:

\(\displaystyle A=-D\)

\(\displaystyle A=D+2\)

Thus, \(\displaystyle A=1,\,D=-1\) and so we find:

[box=green]\(\displaystyle \frac{4x^2y}{\left(x^2-2xy+2y^2\right)\left(x^2+2xy+2y^2\right)}=\frac{x}{x^2-2xy+2y^2}-\frac{x}{x^2+2xy+2y^2}\)[/box]