1
$\begingroup$

enter image description here

I know that I cannot just do $\frac{x^{-1}}{x^{-3}}$ and $\frac{y^{-1}}{y^{-3}}$ and get $x^{2}+y^{2}$ but how can i factor either the numerator or denominator to get something to cancel out?

$\endgroup$
3
$\begingroup$

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

$\endgroup$
2
$\begingroup$

It is $$\frac{\frac{x+y}{xy}}{\frac{x^3+y^3}{x^3y^3}}=\frac{(x+y)x^3y^3}{xy(x^3+y^3)}$$ and now you can use (after cancelling $$xy$$) that $$x^3+y^3=(x+y)(x^2-xy+y^2)$$

$\endgroup$
0
$\begingroup$

A "simpler" expression is

$$\frac1{x^{-2}-x^{-1}y^{-1}+y^{-2}},$$ by factorization.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.