Question

(sqrt х-1)^2=(1-sqrt x)-x(x-1)(x+1)

Answer 1

$(\sqrt{x}-1)^2 = (1-\sqrt{x})-x(x-1)(x+1) \\ x - 2\sqrt{x} + 1 = 1 - \sqrt{x} - x(x^2-1) \\ x - 2\sqrt{x} + 1 = 1 - \sqrt{x} - x^3+x \\ x - 2\sqrt{x} + 1 - 1 + \sqrt{x} + x^3-x = 0 \\ x^3 - \sqrt{x} = 0 \\ x^3 = \sqrt{x} \\ x^6 = x \\ x_1 = 0 \ ; \ x_2 = 1$

Answer 2

Let's solve the equation step by step:

Step 1: Expand the right-hand side (RHS) of the equation. $(1 - \sqrt{x}) - x(x - 1)(x + 1)$

Step 2: Expand the expression $x(x - 1)(x + 1)$: $x(x - 1)(x + 1) = x(x^2 - x - x + 1) = x(x^2 - 2x + 1) = x^3 - 2x^2 + x$

So, the right-hand side (RHS) of the equation becomes: $(1 - \sqrt{x}) - (x^3 - 2x^2 + x) = 1 - \sqrt{x} - x^3 + 2x^2 - x$

Step 3: Square the left-hand side (LHS) of the equation: $(\sqrt{x} - 1)^2 = (\sqrt{x})^2 - 2 \cdot \sqrt{x} \cdot 1 + 1 = x - 2\sqrt{x} + 1$

Now, the equation becomes: $x - 2\sqrt{x} + 1 = 1 - \sqrt{x} - x^3 + 2x^2 - x$

Step 4: Move all terms to one side to obtain a single polynomial equation: $x - 2\sqrt{x} + 1 + \sqrt{x} + x^3 - 2x^2 + x = 0$

Step 5: Combine like terms: $x^3 - 2x^2 + 2x - \sqrt{x} + 1 = 0$

This is the final polynomial equation, and it doesn't seem to have an elementary solution for $x$ that can be expressed explicitly in terms of common mathematical functions.

Depending on the context or the requirements of the problem, you might need to approximate the solution numerically or use numerical methods to find the roots of this equation.