Solve the equation (x^3 - x)^1/2 + (2x - 1)^1/2 = (x^3 + x - 1)1/2.

To solve the equation and to eliminate the square
roots, we'll square raise both sides:

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

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

We'll combine and
eliminate like terms and we'll get:

2sqrt[(x^3 - x)(2x -
1)] = 0

We'll divide by
2:

[(x^3 - x)(2x - 1)] =
0

We'll set each factor as
zero:

x^3 - x = 0

We'll
factorize:

x(x^2-1) = 0

We'll
expand the difference of squares:

x(x-1)(x+1) =
0

We'll set each factor as
zero:

x1 = 0

x-1 =
0

x2 = 1

x+1 =
0

x3 = -1

2x - 1 =
0

2x = 1

x4 =
1/2

Now, we'll check the found results in
equation:

For x1 = 0

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

sqrt(0) + sqrt( -
1)= sqrt( - 1)

But sqrt -1 is
impossible!

For x2 = 1

sqrt(1
-1) + sqrt(2 - 1)= sqrt(1 + 1- 1)

0 + 1 =
1

1=1

So x = 1
is admissible.

For x3 =
-1

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

sqrt(-1 + 1) + sqrt(-2 - 1)= sqrt(-1 -1-
1)

sqrt -3 = sqrt-3

But,
sqrt-3 is impossible!

For x =
1/2

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

sqrt(1/8- 1/2) + sqrt(2/2 - 1)= sqrt(1/8 +1/2 -
1)

But sqrt -3/8 is
impossible!

So, the only admissible solution
of the equation is x = 1.