# Thread: Need help with exponents: simplifying 2

1. ## Need help with exponents: simplifying 2

One more question. Is (x^-1 + y^-1)^-2 = x^2 + y^2? Thanks anyone who can help, please answer =)

2. Originally Posted by view360
One more question. Is (x^-1 + y^-1)^-2 = x^2 + y^2? Thanks anyone who can help, please answer =)

you've got that one right.

3. Originally Posted by view360
One more question. Is (x^-1 + y^-1)^-2 = x^2 + y^2? Thanks anyone who can help, please answer =)
Originally Posted by princess_21
you've got that one right.
also, this is not right. we can distribute powers across products, not sums

$(a + b)^2 \ne a^2 + b^2$ for instance, but rather $(a + b)^2 = (a + b)(a + b) = a^2 + 2ab + b^2$

similarly, $(x^{-1} + y^{-1})^{-2} = \frac 1{(x^{-1} + y^{-1})^2} = \frac 1{x^{-2} + 2(xy)^{-1} + y^{-2}}$

if you don't like negative powers in the denominator, you could approach it this way

$(x^{-1} + y^{-1})^{-2} = \left( \frac 1x + \frac 1y \right)^{-2} = \left( \frac {x + y}{xy} \right)^{-2} = \frac {x^2y^2}{(x + y)^2} = \frac {x^2y^2}{x^2 + 2xy + y^2}$

also, please post new problems in a new thread. it saves me the work of having to split the thread up

4. Originally Posted by Jhevon
also, this is not right. we can distribute powers across products, not sums

$(a + b)^2 \ne a^2 + b^2$ for instance, but rather $(a + b)^2 = (a + b)(a + b) = a^2 + 2ab + b^2$

similarly, $(x^{-1} + y^{-1})^{-2} = \frac 1{(x^{-1} + y^{-1})^2} = \frac 1{x^{-2} + 2(xy)^{-1} + y^{-2}}$

if you don't like negative powers in the denominator, you could approach it this way

$(x^{-1} + y^{-1})^{-2} = \left( \frac 1x + \frac 1y \right)^{-2} = \left( \frac {x + y}{xy} \right)^{-2} = \frac {x^2y^2}{(x + y)^2} = \frac {x^2y^2}{x^2 + 2xy + y^2}$

also, please post new problems in a new thread. it saves me the work of having to split the thread up
sorry again, my mistake. thanks Jhevon.

5. Originally Posted by princess_21
sorry again, my mistake. thanks Jhevon.
don't worry about it, princess, it happens to the best of us. would you believe i've made worse errors here before? the good thing is, you always get feedback if you mess up, someone catches you. it's a team and it's a pleasure to have you help out

6. thank you again. without u correcting me i would have given a wrong answer. that is what i like here in this forum there are always members who can help ..