HELP! PLEASE!:Find the inverse and verify using composition

1)  (2x+3)^3 -6

Write

y = (2x+3)^3 - 6     add 6 to both sides

y + 6 =  (2x+ 3)^3      take the cube root of each side

∛ (y + 6)  = 2x + 3      subtract 3 from both sides

∛ ( y + 6) - 3  =  2x      divide both sides by 2

 [  ∛  ( y + 6) - 3 ] / 2   = x     "swap"  x and y

[ ∛ ( x + 6) - 3 ] / 2   = y  = the inverse

Putting this into    (2x+3)^3 -6  we get

( 2  [ ∛ ( x + 6) - 3 ] / 2  + 3)^3 - 6  =

(  ∛ ( x + 6) - 3 + 3 )^3  -  6 =

x + 6  - 6  = x

And    putting the original into the inverse

[ ∛ ( [   (2x+3)^3 -6 ] + 6) - 3 ] / 2  =

[  ∛ { (2x + 3)^3 - 3 ] / 2 =

[2x + 3 - 3 ] / 2  =

2x / 2 =   x

So...they are inverses !!!!

2) 

y =  (-3x+1)^2 - 2

y + 2 =  (-3x + 1)^2       take the root

sqrt ( y + 2) = -3x + 1

sqrt (y + 2) - 1 =  -3x

[ 1 - sqrt (y + 2) ] / 3  =  x           swap   x and y

[ 1 - sqrt (x + 2) ] / 3 =  y         this is the inverse

Put the original into this 

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

[ 1 - sqrt [ -3x  +1 }^2  ] / 3

[1 - [ -3x + 1 ] ] / 3

3x /3     =  x

Put the inverse into the original

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

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

( sqrt (x + 2) )^2 - 2  =  

x + 2 - 2   =   x

3)  square root of (3x+7)

y = √[3x + 7]        square both sides

y^2 = 3x + 7

y^2 - 7   =  3x

[ y^2 - 7 ] /3 = x

[ x^2 - 7 ] / 3 = y  = the inverse

Put the inverse into the original

√ (  3 [ [ x^2 - 7 ] / 3  + 7 )  =

√  [ x^2 - 7 + 7 ] =

√x^2   = x

Put the original into the inverse

[ ( √[3x + 7]  ) ^2 - 7 ] / 3

[ 3x + 7 - 7 ) /3

3x /3 =   x

y =  -2x^3 +6

y - 6 = -2x^3

[ 6 - y ] / 2 = x^3

∛ ( [ 6 - y ] / 2 )  = x

∛ ( [ 6 - x] / 2) = y = the inverse

Original into inverse

∛ ( [ 6 - (    -2x^3 +  6     )   ] / 2)

∛ ( [ 2x^3 ] / 2 ) =

∛ x^3 = x

Inverse into original

-2(  ∛ ( [ 6 - x] / 2)   )^3 +  6  =

-  ∛ [ 6 - x]^3  + 6  =

- 6 + x + 6    =    x