How to get correct partial fraction decomposition of this expression?
$$\frac{3x+3y-z}{(x+y+z)^3}$$
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this is what i have found $$3, \left( x+y+z \right) ^{-2}-4,{\frac {z}{ \left( x+y+z \right) ^{3 }}} $$ – Dr. Sonnhard Graubner Jul 25 '17 at 11:38
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Hint
$$\frac{A}{x+y+z}+\frac{B}{(x+y+z)^2}+\frac{C}{(x+y+z)^3}$$
Now find $A,B,C$.
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$\frac{A}{x+y+z} + \frac{B}{(x+y+z)^2} + \frac{B}{(x+y+z)^2} = \frac{3x+3y-z}{(x+y+z)^3}$
multiply both sides by $(x+y+z)^3$
$A(x+y+z)^2 + B(x+y+z) + C = 3x+3y-z$ $A(x^2+y^2+z^2+2xy +2xz +2yz) + Bx+By+Bz + C = 3x+3y-z$
compare both sides of equation $A=0$, $B= 3$ according to comparison $x$ and $y$ or $B= -1$ according to comparison with $z$, what about the C?
– Vid Jul 25 '17 at 11:56 -
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