Let `2^x = 3^y = 6^-z = k`    `hArr  2 =k ^(1/x) ,  3 = k^(1/y)  and  6 =  K^-(1/z)`

Now  2 x 3  =   `hArr      k^(1/x) xx k^(1/y) =  k^-(1/z)         hArr    k^(1/x+1/y)  =  k^-(1/z)`

`:.`    `1/x + 1/y =  - 1/z     hArr    1/x + 1/y + 1/z = `o.

`a^x = b^y = c^z   = k`. Then`, a = k^(1/x) , b = k^(1/y) and c =  k^(1/z)`.

`:.`    `b^2 =  ac   hArr   (k^(1/y))^2 =  k^(1/x) xx k^(1/z)        hArr   k^(2/y) =   k  ^(1/x + 1/z)`

`:.`      `2/y = ((x + z))/(xz)   hArr     y/2     =  (xz)/((x + z))     hArr   y =   (2xz)/((x + z))`

`5^(x + 3) = (25)^(3x - 4)`   `hArr  5^(x + 3) = (5^2)^(3x - 4)           hArr   5^(x + 3)    =  5^(2(3x-4))`

`hArr  5^(x + 3)  = 5^(6x - 8)     hArr x + 3 =   6x - 8         hArr  5x = 11         hArr     =   11/5`

`2^x xx 8^(1/5) =  2^(1/5)   hArr   2^x xx (2^3)^(1/5)   = 2^(1/5)     hArr  2^x xx 2^(3/5) =  2^(1/5)`

`hArr    2^(x + 3/5) = 2^(1/5)`

`hArr   x + 3/5 =  1/5  hArr   x = (1/5 - 3/5) =  -(2)/(5).`

`2^x = root3(32)`  `hArr  2^x  = (32)^(1/3)  =  (2^5)^(1/3)= 2^(5/3)   hArr   x =  5/3`.

`(9/4)^x.(8/27)^(x - 1)  =  2/3`    `hArr`   ` 9^x/4^x xx  8^(x - 1)/(27)^(x - 1) = 2/3`

`hArr (3^2)^x/(2^2)^x xx((2^3)^(x - 1))/((3^3)^(x - 1))  =  2/3     hArr   (3^(2x) xx 2^(3(x- 1)))/(2^(2x) xx 3^(3(x - 1))) = 2/3`

`hArr  (2^((3x - 3 -2x)))/(3^((3x - 3 - 2x))) = 2/3     hArr  (2^((x-3)))/(3^((x- 3)))  = 2/3     hArr (2/3)^(x -3)`

=`(2/3)^1   hArr  x - 3 = 1    hArr  x  = 4.`

= `3^(x - y) = 3^3      hArr     x - y = 3`..........................(i)

= `3^(x + y) = 3^5     hArr     x + y =  5` .........................(ii)

On solving (i) and (ii) , we  get   `x` = 4.

`sqrt(a^-1 b), sqrt(b^-1 c), sqrt(c^-1 a) `  =`(a^-1)^(1/2).b^(1/2).(b^-1)^(1/2). c^(1/2).(c^-1)^(1/2).a^(1/2)`

=`(a^-1 a)^(1/2).(b.b^-1)^(1/2).(c.c^-1)^(1/2)` = `(1)^(1/2).(1)^(1/2).(1)^(1/2)` = (1 x 1 x 1 ) = 1.

Given Exp. `(1)/(1 + a + b^-1) + (1)/(1 + b + c^-1) + (1)/(1 + c + a^-1)`

=  `(1)/(1 + a + b^-1) + (b^-1)/(b^-1+1 + b^-1  c^-1) + (a)/(a + ac + 1)`

= `(1)/(1 + a + b^-1) + (b^-1)/(1 + b^-1 + a) + (a)/(a + b^-1 + 1)`

=`(1 + a + b^-1)/(1 + a + b^-1)` = 1.

[  `:.`   abc = 1   `rArr`    `(bc)^-1`  = a    `rArr`    `b^-1 c^-1`  =  a  and  ac = `b^-1` ]

Given Exp.  `[x^(a - b)]^(1/(ab))`.`[x^(b - c)]^(1/(bc))`.  `[x^-(c - a)]^(1/((ca))`

=`x^(((a - b)) /(ab))`.`x^(((b - c))/(bc))`. `x^(((c - a))/(ca))`

= `x^{{((a - b))/(ab) + ((b - c))/(bc)  + ((c - a))/(ca)}}`

=`x^((1/b -1/a) + (1/c - 1/b) + (1/a - 1/c))`         = `x^0`    = 1.