Sol.         `(x^-1 - 1)/(x - 1)`

              = `(1/x - 1)/(x - 1)`

               = `(1 - x)/x xx 1/(x - 1)`

                = `- 1/x`

Hence , the required quotient is `- 1/x`

Sol.        `(256)^0.18  xx (16)^ 0.18`

              = `[(16^2)^0.16 xx (16)^0.16]`

              = `(16)^(2 xx 0.16) xx (16)^0.18`

             = `(16)^0.32 xx (16)^0.18`

            = `(16)^(0.32 + 0.18)`

            = `(16)^0.5`       = `(16)^(1/2) = 4`  

Sol .         `(.00032)^(3/5)`

                 = `(32/100000)^(3/5)`

                  =`(2^5/10^5)^(3/5)`

                  = `{(2/10)^5}^(3/5)`    = `(1/5)^(5 xx 3/5)` = `(1/5)^3` = `1/125`

Sol .     `(8/125)^-(4/3)`

             = `{(2/5)^3}^-(4/3)`

             = `(2/5)^{ 3 xx (-4)/3}`

             = `(2/5)^-4`

            = `(5/2)^4`   = `5^4/2^4`          = `625/16` 

Sol .       `(1024)^-(4/5)`

               = `(4^5)^-(4/5)`

               = `4^{5 xx (-1)/5}`

                = `4^-4`    = `1/4^4`

                 = `1/256` 

Sol.      `(27)^(2/3)` 

            = `(3^3)^(2/3)`

              = `3^(3 xx 2/3)`

              = `3^2` = 9. 

`x = 5 + 2sqrt(6)  = 3 + 2 + 2sqrt(6) =  (sqrt(3))^2 + (sqrt(2))^2 + 2 xx sqrt(3)xxsqrt(2) = (sqrt(3) + sqrt(2))^2`

Also `, (x - 1) =  4 + 2sqrt(6)  = 2(2 + sqrt(6)) =  2sqrt(2)(sqrt(2) + sqrt(3))`

`:.`  `  (x - 1)/sqrt(x)  =  (2sqrt(2)(sqrt(3) + sqrt(2)))/((sqrt(3) + sqrt(2))) = 2sqrt(2)`.

L.C.M. of 2, 3, 4 is 12

`sqrt(2) = 2^(1/2) = 2^((1/2 xx 6/6)) =  2^(6/12) = (2^6)^(1/12) =  (64)^(1/12) =   root12(64)`

`root3(3) = 3^(1/3) =  3^((1/3 xx 4/4)) =  3^(4/12) =  (3^4)^(1/12) =  (81)^(1/12) =  root12(81)`

`root4(4) = 4^(1/4) =  4^((1/4 xx 3/3)) = 4^(3/12) =  (4^3)^(1/12 )=  (64)^(1/12) =  root12(64)`

Clearly, `root12(81),  i.e. , root3(3) ` is the largest.

`2^x = 4^y  = 8^z     hArr  2^x = 2^(2y) = 2^(3z)     hArr   x = 2y = 3z`

`:.    (1)/(2x) + (1)/(4y) + (1)/(6z) =  24/7   hArr  (1)/(6z) + (1)/(6z) + (1)/(6z) =  24/7`

`hArr (3)/(6z) = 24/7    hArr   z = (3/6 xx 7/24)  =  7/48`

`a^1 =  c^z =  (b^y)^z =      b^(yz) = (a^x)^(yz) =   a^(xyz)     :.     xyz =  1. `