Question

http://prntscr.com/40fxbu help me prove this two expression are the same

Answer 1

So you want to simplify the left side?

Answer 2

yeah basically the right side is the answer

Answer 3

please help i kind of stuck

Answer 4

All you need to know is,

\[\huge x ^{-n} = \frac{ 1 }{ x }\]

Answer 5

i know that let me show you how im doing i just cant seem to get to the answer

Answer 6

Alright

Answer 7

\[\huge \frac{ a ^{-1}b ^{-2}+a ^{-2}b ^{-1} }{ b ^{-2}-a ^{-2} }\]

I'll write it out here for reference.

Answer 8

http://prntscr.com/40ghj3

Answer 9

so thats where im stuck

Answer 10

ok let me check

Answer 11

Ok sec, let me do this, just need to put it over a common denominator I guess.

Answer 12

Guess it will require some work haha, alright I'll do this and you can see my steps alright?

Answer 13

ok nice, beacause i cant seem to find things to cancel out

Answer 14

ok thank you.

Answer 15

\[\huge \frac{ a ^{-1}b ^{-2}+a ^{-2}b ^{-1} }{ b ^{-2}-a ^{-2} } = \frac{ \frac{ 1 }{ ab^2 }+\frac{ 1 }{ a^2b } }{b ^{-2}-a ^{-2} }\]

Agree?

Answer 16

yeah

Answer 17

Ok now we'll put it over a common denominator

\[\huge \frac{ 1 }{ ab^2 }+\frac{ 1 }{ a^2b } = \frac{ a }{ a^2b^2 }+\frac{ b }{ a^2b^2 } \implies \frac{ \frac{ a }{ a^2b^2 }+\frac{ b }{ a^2b^2 } }{ b ^{-2}-a ^{-2} }\]

Answer 18

oj i cant see part of it

Answer 19

Yeah sorry, I lost connection OS, you know how it is >_>

Answer 20

hehe

Answer 21

\[\huge \frac{ \frac{ a }{ a^2b^2 }+\frac{ b }{ a^2b^2 } }{ b ^{-2}-a ^{-2} }\]

\[\huge = \frac{ \frac{ a+b }{ a^2+b^2 } }{ b ^{-2}-a ^{-2} }\]

\[\huge = \frac{ a+b }{ a^2b^2\frac{ a^2 }{ a^2b^2 }-\frac{ b^2 }{ a^2b^2 } }\]

put b^-2-a^-2 over the common denominator

\[\huge = \frac{ a+b }{ a^2b^2\frac{ a^2-b^2 }{ a^2b^2 } } \]

\[\huge = \frac{ a+b }{ a^2b^2 \frac{ (a-b)(a+b) }{ a^2b^2 } }\]

Using difference of squares

\[\huge = \frac{ a^2b^2(a+b) }{ a^2b^2(a-b)(a+b) } \implies \frac{ 1 }{ a-b }\]

cancelling common terms.

Answer 22

whoa let me digest that please.

Answer 23

Take your time ^.^

Answer 24

my friend did you sum again the fraction right to obtein a+b / a^2+b^2 right?

Answer 25

http://prntscr.com/40goxu could you explain me the procedure to get that