SOLUTION: Power reduce sin^3⁡x as a sum (or difference) of first degree trig terms

The trick is to start with sin(3x) and simplify
and it will come out with an equation that you
can solve for  sin³(x).

sin(3x) = sin(2x+x) = sin(2x)cos(x)+cos(2x)sin(x) = 

2sin(x)cos(x)cos(x)+cos(2x)sin(x) =

2sin(x)cos²(x)+[1-2sin²(x)]sin(x) =

2sin(x)[1-sin²(x)]+[1-2sin²(x)]sin(x) =
  
2sin(x) - 2sin³(x) + sin(x) - 2sin³(x) =

3sin(x) - 4sin³(x)

So

sin(3x) = 3sin(x) - 4sin³(x)

4sin³(x) = 3sin(x) - sin(3x)

sin³(x) = sin(x) - sin(3x)

Edwin