Determine the area of the figure bounded by two branches of the curve `(y-x)^(2)=x^(3)` and the straight line `x=1`.

Answer

To determine the area of the figure bounded by the two branches of the curve \((y - x)^2 = x^3\) and the straight line \(x = 1\), we can follow these steps:

### Step 1: Rewrite the curve equation
The given equation is \((y - x)^2 = x^3\). We can express \(y\) in terms of \(x\):
\[ y - x = \pm x^{3/2} \]
This gives us two equations:
1. \(y = x + x^{3/2}\) (let's call this \(f_1\))
2. \(y = x - x^{3/2}\) (let's call this \(f_2\))

### Step 2: Identify the area to be calculated
We need to find the area between the two curves \(f_1\) and \(f_2\) from \(x = 0\) to \(x = 1\).

### Step 3: Set up the integral for the area
The area \(A\) between the curves can be calculated using the integral:
\[ A = \int_0^1 (f_1 - f_2) \, dx \]
Substituting \(f_1\) and \(f_2\):
\[ A = \int_0^1 \left((x + x^{3/2}) - (x - x^{3/2})\right) \, dx \]
This simplifies to:
\[ A = \int_0^1 (2x^{3/2}) \, dx \]

### Step 4: Calculate the integral
Now we calculate the integral:
\[ A = 2 \int_0^1 x^{3/2} \, dx \]
Using the power rule for integration:
\[ \int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C \]
For \(n = \frac{3}{2}\):
\[ \int x^{3/2} \, dx = \frac{x^{5/2}}{5/2} = \frac{2}{5} x^{5/2} \]
Now, evaluate from 0 to 1:
\[ A = 2 \left[ \frac{2}{5} x^{5/2} \right]_0^1 = 2 \left( \frac{2}{5} (1) - 0 \right) = 2 \cdot \frac{2}{5} = \frac{4}{5} \]

### Final Answer
The area of the figure bounded by the two branches of the curve and the line \(x = 1\) is:
\[ \boxed{\frac{4}{5}} \text{ square units} \]