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

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

=`x^((a^2 - b^2  + b^2 -c^2 + c^2 - a^2))`      =`x^0` = 1.

Given Exp.= `x^((b - c)(b + c - a))` .          `x^((c - a)(c + a - b))`.           `x^((a - b)(a + b - c))`

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

=`x^((b^2 -c^2 + c^2 -a^2 + a^2 - b^2))`  .  `x^ - a(b - c) - b(c - a) -c(a - b)`        =`(x^0 xx x^0)` =  1 x 1 = 1.

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

= `(x^a)/((x^a + x^b + x^c))` +` (x^b)/((x^a + x^b + x^c))` +` (x^c)/((x^a + x^b + x^c))`

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

`(1)/(1 + a^((n -m)))+ (1)/(1 + a^((m - n)))`  = `(1)/((1 + a^n/a^m))+ (1)/((1 + a^m/a^n))`

=`(a^m)/((a^m + a^n)) + (a^n)/((a^m + a^n))`    = `((a^m + a^n))/((a^m + a^n))` =  1.

`(6^12 xx (35)^28 xx (15)^16)/((14)^12 xx (21)^11)`

=`((2 xx 3)^12 xx (5 xx 7)^28 xx (3 xx 5)^16)/((2 xx 7)^12 xx (3 xx 7)^11)`

=`(2^12 xx 3^12 xx 5^28 xx 7^28 xx 3^16 xx 5^16)/(2^12 xx 7^12 xx 3^11 xx 7^11)`

=`2^(12 - 12) xx 3^(12 + 16 - 11) xx 5^(28 + 16) xx 7^(28 - 12 - 11)`

=`2^0 xx 3^17 xx 5^44 xx 7^ -5` =  `(3^17 xx 5^44)/(7^5)`

Number of prime factors = 17 + 44 + 5 = 66.

`(216)^(3/5) xx (2500)^(2/5) xx (300)^(1/5)`  = `(3^3 xx 2^3)^(3/5)  xx (5^4 xx 2^2)^(2/5) xx (5^2 xx 2^2  xx3)^(1/5)`

=`3^((3xx3/5)) xx 2^((3xx3/5)) xx 5^((4xx2/5)) xx 2^ ((2xx2/5)) xx 5^((2xx1/5)) xx 2^((2xx1/5)) xx 3^(1/5)`

=`3^(9/5) xx 2^(9/5) xx5^(8/5) xx 2^(4/5) xx 5^(2/5) xx 2^(2/5) xx 3^(1/5)`

=`3^((9/5 + 1/5)) xx 2^((9/5 + 4/5 + 2/5)) xx 5^((8/5 + 2/5))`

=`3^2 xx 2^3 xx 5^2`.          

Hence, the number of prime factors = (2 + 3 + 2 ) = 7.

`((243)^(n/5) xx 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(5xxn/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^(2n -1))` = `3^(3n + 1 - 3n + 1)`= `3^2` = 9.

We  know that  `11^2` = 121 , putting  m = 11 and  n = 2, we get :

`(m - 1)^(n + 1) =  (11 - 1)^(2 + 1) =   10^3`  = 1000.

`x^z = y^2   hArr  (10^0.48)^z =  (10^0.70)^2     hArr  10^(0.48z)   =  10^(2 xx -.70)  = 10^1.40`

         `hArr`  0.48z =  1.40    ` hArr    z = 140/48          =  35/12 `    = 2.9 (approx.)

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

=`(3 + 2sqrt(2)) + 1/((3 + 2sqrt(2))) xx ((3 - 2sqrt(2)))/((3 - 2sqrt(2))) - 2 = (3 + 2sqrt(2)) + (3 - 2sqrt(2)) - 2  = 4`

`:.`   `(sqrt(x) -  1/sqrt(x))` = 2.