Sol.      Given surds are of order 2 and 3.  Their L.C.M. is 6

             Changing each to a surd of order 6, we get :

              `sqrt(2) =  2^(1/2) = 2^(1/2 xx 3/3) = (2^3)^(1/6) = (8)^(1/6) = root6(8)`

               `3sqrt(3) = 3^(1/3) = 3^(1/3 xx 2/2) = 3^(2/6) = (3^2)^(1/6)= root6(9)`

Clearly `, root6(9)  > root6(8) `  and  hence `root3(3) > sqrt(2)`

Sol.     Given expression.

=`x^((a - b)(a^2 + b^2 + ab))xx  x^((b - c)(b^2 + c^2 + bc)) xx  x^((c - a)(c^2 + a^2 + ca))`

=`x^(a^3 - b^3) x^(b^3 - c^3) x^(c^3 - a^3)`

 =` x^(a^3 - b^3 + b^3 - c^3 + c^3 - a^3)`

= `x^0` = 1.

Sol .        `z^1 =  x^c = (y^a)^c         because                          [        x = y^a]`

                = ` y^(ac) = (z^b)^(ac)     because                            [         y = z^b]`              

                = `z^(b(ac)) = z^(abc)` 

`:.`             abc = 1. 

Sol .       `(6^(2/3) xx root3(6^7))/(root3(6^6))`

               = `(6^(2/3) xx (6^7)^(1/3))/(6^6)^(1/3)`

                = `( 6^(2/3) xx 6^(7 xx 1/3))/(6^(6 xx 1/3))`

               = `(6^(2/3)  xx 6^( 7/3))/(6^2)` 

                = `6^(2/3) xx 6^(7/3 - 2)`

                = `6^(2/3) xx 6^(1/3)`

                 = ` 6^((2/3 + 1/3))` =  `6^1 = 6`

Sol.      Putting   `2^(1/4) = x`, we get :

            `(2^(1/4) - 1) ( 2^(3/4) + 2^ (1/2) + 2^(1/4) + 1)`

              = `( x - 1) ( x^3 + x^2 + x + 1)`  where  `x = 2^(1/4)`

              = ` (x - 1) {x^2( x + 1) + (x + 1)}`

               = `(x - 1) ( x + 1 ) (x^2 + 1)`

              = `(x^2 - 1) (x^2 + 1)`

             = `(x^4 -  1)`

             = `[2^((1/4)^4) - 1]`           =` [2 ^(1/4 xx 4) - 1  ]   = 2 - 1 = 1`.  

Sol .   `((243)^(n/3) . 3^(2n + 1))/(9^n xx 3^(n - 1))`

           = `((3^5)^(n/5) xx 3^(2n + 1))/((3^2)^(n) xx 3^(n - 1))`

            = `(3^(5 xx n/5) xx 3^(2n + 1))/(3^(2n) xx 3^(n - 1))`

            =`(3^n xx 3^(2n + 1))/(3^(2n) xx 3^(n - 1))`

           = ` (3^(n + (2n + 1)))/(3^(2n + n - 1)) `

            = `(3^(3n + 1))/(3^(3n - 1))`

            = `3^((3n + 1) - (3n - 1)) = 3^2` = 9.  

Sol.        `(1/5)^3y = 0.008`   = `8/1000 =  1/125 = (1/5)^3`

                `(1/5)^3y  = (1/5)^3`

`hArr           3y = 3`

`hArr             y = 1.`

`:.`          `   (0.25)^y = (0.25)^1     = 0.25`

Sol .      `[(16)^(3/2) + (16)^(- 3/2)`

              =  `[(4^2)^(3/2) + (4^2)^(- 3/2)`

               = ` 4^(2 xx 3/2) + 4^( 2 xx (-3)/2)`

               =` 4^3 + 4^-3`

               = `4^3 + 1/4^3`   =` 64 + 1/64`

                 =` 4097/64`

Sol.    `[5(8^(1/3) + 27^(1/3))^3]^(1/4)`

          = `[5(2^3)^(1/3) + (3^3)^(1/3))^3]^(1/4)`

          =`[5{2^(3 xx 1/3) + 3^(3 xx 1/3)}^3]^(1/4)`

          = `[5(2 + 3)^3]^(1/4)`

           = `(5 xx 5^3)^(1/4)`

           = `(5^4)^(1/4)` = `5^(4 xx 1/4)` = `5^1`  = 5.

Sol.          `2^(x - 1) + 2^(x + 1)` = 1280

`hArr         2^(x  -1) (1 + 2^2) = 1280`

 `hArr        2^(x - 1) = 1280/5`

`hArr        2^(x - 1) = 256  = 2^8`

`hArr        x - 1 = 8`

`hArr         x = 9`

Hence  x = 9.