How can one solve the equation $\sqrt{x\sqrt{x} - x} = 1-x$?

For the left hand side to be defined, you need $x\geq 1$ or $x=0$. Zero is not a solution. The left hand side will equal 0 for $x=1$ and will be strictly positive if $x>1$. For $x\geq 1$, the right hand side is equal to 0 if $x=1$ and will be strictly negative if $x>1$. This shows that the only real solution is $x=1$.

If $\sqrt{x\sqrt{x}-x} = 1-x$, squaring both sides we have $x \sqrt{x} - x = (1-x)^2$, and then taking the $-x$ to the other side and squaring again we get $x^3 = (x + (1-x)^2)^2$. Simplify to $(x-1)(x^3 -2x^2 +x-1) = 0$, the second factor being irreducible over the rationals. Of course $x=1$ is a solution. The roots of the cubic are rather complicated. One is real (approximately $1.754877666$), but it is not a solution of the original equation because the right side would be negative and the square root of a positive number is positive. For the complex roots, you have to specify which branch of the square root you mean. If you mean the principal branch (i.e. nonnegative real part), the complex roots are also not solutions of the original equation.

The equation $$\sqrt{x\sqrt{x}-x} = 1-x$$ implies that both sides, and the argument of each square root, are all nonnegative. Thus any solution must obey $$\sqrt{x}\text{ exists}\implies x\ge0$$ $$0\le1-x\implies x\le1$$ $$0\le x\sqrt{x}-x=x\left(\sqrt{x}-1\right) \implies \sqrt{x}\ge1 \implies x\ge1$$ But then $1\le x\le 1 \implies x=1$ is the only solution.

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 $x=1$.