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If x=a+a/r+a/(r^2)+oo,y=b-b/r+b/(r^2)+oo...

If `x=a+a/r+a/(r^2)+oo,y=b-b/r+b/(r^2)+oo,a n dz=c+c/(r^2)+c/(r^4)+oo,` prove that `(x y)/z=(a b)/cdot`

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We have,
`x=a/(1-1/r)=ar/(r-1)`
`y=b/(1-(-1/r))=(br)/(1+r)`
`z=c/(1-1/r^(2))`
`thereforexy=((ar)/(r-1))((br)/(r+1))=(abr^(2))/(r^(2)-1)`
`therefore (xy)/z=[((abr^(2))/(r^(2)-1))/((cr^(2))/(r^(2)-1))]=(ab)/c`
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