Rational Expressions - Complex functions SOS!!!

**edmundyong** · November 27th 2012, 01:10 AM

(x^-3+ y^-3) / (x^-2 – x^-1 y^-1 + y^-2)

the answer is (x + y) / xy .

p.s '/' indicated fraction.

I tried it for almost 45 minutes but stuck at the last part.
I get this : [x+y (x²-xy-y²)] / [xy (x²+y²)]
This is my first post so please prove this forum worth staying !!!

**MarkFL** · November 27th 2012, 01:21 AM

This is how I would simplify:

$\frac{x^{-3}+y^{-3}}{x^{-2}-x^{-1}y^{-1}+y^{-2}}\cdot\frac{x^{3}y^{3}}{x^{3}y^{3}}=$

$\frac{y^3+x^3}{xy(y^2-xy+x^2)}=\frac{(x+y)(x^2-xy+y^2)}{xy(x^2-xy+y^2)}=\frac{x+y}{xy}$

Another approach:

$\frac{x^{-3}+y^{-3}}{x^{-2}-x^{-1}y^{-1}+y^{-2}}=\frac{(x^{-1}+y^{-1})(x^{-2}-x^{-1}y^{-1}+y^{-2})}{x^{-2}-x^{-1}y^{-1}+y^{-2}}=$

$\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}$

**Prove It** · November 27th 2012, 01:22 AM

$\displaystyle \begin{align*} \frac{x^{-3} + y^{-3}}{x^{-2} - x^{-1}y^{-1} + y^{-2}} &= \frac{\frac{1}{x^3} + \frac{1}{y^3}}{\frac{1}{x^2} - \frac{1}{xy} + \frac{1}{y^2}} \\ &= \frac{\frac{x^3 + y^3}{x^3y^3}}{\frac{y^2 - xy + x^2 }{x^2y^2}} \\ &= \frac{x^2y^2 \left( x^3 + y^3 \right)}{x^3y^3 \left( y^2 - xy + x^2 \right)} \\ &= \frac{x^3 + y^3}{xy \left( y^2 - xy + x^2 \right)} \\ &= \frac{(x + y) \left( x^2 - xy + y^2 \right)}{ xy \left( y^2 - xy + x^2 \right)} \\ &= \frac{x + y}{xy} \end{align*}$

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