Evaluate :

`(1 - 1/3 )^2 times ( 1 - 1/4 )^2 times ( 1 - 1/5 )^2 ------- ( 1 - 1/n )^2` is equal to :

Evaluate : ( 1 - 1/2 ) times ( 1 - 1/3 ) times ( 1 - 1/4 ) times ( 1 - 1/5 ) =

Evaluate : [ ( -1/3 )^2 times ( -1/3 )^3 ] ÷ [ ( -1/6 )^2 times ( -1/6 )^3 ]

Evaluate : (-3)^2 times ( 1/3 )^5 times (1/2 )^4 times (-4 )^2

Evaluate : ( 1/2 ÷ 1/2 times 1/2 ) / ( 1/2 + 1/2 times 1/2 )

Solve : 1/2 times ( 1/3 - 1/4 ) = 1/3 times ----- - 1/2 ------

Evaluate : 4^(1/3 ) times [ 2^(1/3 ) times 3^(1/2) ] ÷ 9^(1/4 )

Evaluate : 1/2 times 1/6 + 1/2 times 1/6 ÷ 1/6 times 1/3

Evaluate : ( 1 + 1/2 ) times ( 1 + 1/3 ) times ( 1 + 1/4 ) times ( 1 + 1/5 ) times ( 1 + 1/6 ) times ( 1 + 1/7) times ( 1 + 1/8 ) is equal to

Evaluate : ( 3/2 )^-1 times ( 3/2 )^-1 times ( 3/2 )^-1 times ( 3/2 )^-1

Evaluate : (-1/2) times (-1/2 ) times ( -1/2 ) times (-1/2) times (-1/2 ) times ( -1/2 )