How can one solve the equation $\sqrt{x\sqrt{x} - x} = 1-x$? $$\sqrt{x\sqrt{x} - x} = 1-x$$
I know the solution but have no idea how to solve it analytically.

Just writing out Robert's manipulation:
$$\eqalign{
  & \sqrt {x\sqrt x  - x}  = 1 - x  \cr 
  & x\sqrt x  - x = {\left( {1 - x} \right)^2}  \cr 
  & x\left( {\sqrt x  - 1} \right) = {\left( {1 - x} \right)^2}  \cr 
  & \sqrt x  - 1 = \frac{{{{\left( {1 - x} \right)}^2}}}{x}  \cr 
  & \sqrt x  = \frac{{{{\left( {1 - x} \right)}^2}}}{x} + 1  \cr 
  & x = {\left( {\frac{{{{\left( {1 - x} \right)}^2}}}{x} + 1} \right)^2}  \cr 
  & x = {\left( {\frac{{1 - 2x + {x^2}}}{x} + 1} \right)^2}  \cr 
  & x = {\left( {\frac{1}{x} + x - 1} \right)^2}  \cr 
  & x = {x^2} - 2x + 3 - \frac{2}{x} + \frac{1}{{{x^2}}}  \cr 
  & {x^3} = {x^4} - 2{x^3} + 3{x^2} - 2x + 1  \cr 
  & 0 = {x^4} - 3{x^3} + 3{x^2} - 2x + 1  \cr 
  & 0 = \left( {x - 1} \right)\left( {{x^3} - 2{x^2} + x + 1} \right) \cr} $$
Note you will most probably have two complex solutions apart from $x=1$.
