For the system of equation
`"log"_(10) (x^(3)-x^(2)) = "log"_(5)y^(2)`
`"log"_(10)(y^(3)-y^(2)) = "log"_(5) z^(2)`
`"log"_(10)(z^(3)-z^(2)) = "log"_(5)x^(2)`
Which of the following is/are true?

Options

A. there are infinite number of solutions
B. there is unique solution with x,y, z `in `Q
C. there are exactly two solutions with x, y, z `in `Q
D. there is no solution

Explanation

To solve the given system of equations, we start with the equations:

1. \(\log_{10}(x^3 - x^2) = \log_{5}(y^2)\)
2. \(\log_{10}(y^3 - y^2) = \log_{5}(z^2)\)
3. \(\log_{10}(z^3 - z^2) = \log_{5}(x^2)\)

### Step 1: Rewrite the equations using properties of logarithms
From the first equation, we can express it in exponential form:
\[ x^3 - x^2 = 5^{\log_{5}(y^2)} = y^2 \]
Thus, we have:
\[ x^3 - x^2 = y^2 \quad \text{(1)} \]

From the second equation:
\[ y^3 - y^2 = z^2 \quad \text{(2)} \]

From the third equation:
\[ z^3 - z^2 = x^2 \quad \text{(3)} \]

### Step 2: Analyze the function \(f(t) = t^3 - t^2\)
We can define a function:
\[ f(t) = t^3 - t^2 = t^2(t - 1) \]
This function is defined for \(t > 1\) and is increasing for \(t > 1\).

### Step 3: Establish relationships between \(x\), \(y\), and \(z\)
Since \(f(t)\) is increasing, if \(x > y\), then \(f(x) > f(y)\), which implies:
\[ y^2 > z^2 \implies y > z \quad \text{(from (1) and (2))} \]

Continuing this reasoning:
- If \(y > z\), then \(f(y) > f(z)\) implies:
\[ z^2 > x^2 \implies z > x \quad \text{(from (2) and (3))} \]
- If \(z > x\), then \(f(z) > f(x)\) implies:
\[ x^2 > y^2 \implies x > y \quad \text{(from (3) and (1))} \]

This leads to a contradiction, suggesting that \(x\), \(y\), and \(z\) must be equal.

### Step 4: Conclude that \(x = y = z\)
Since \(x\), \(y\), and \(z\) cannot be greater than or less than each other, we conclude:
\[ x = y = z \]

### Step 5: Solve for the common value
Substituting \(y = x\) into the first equation:
\[ \log_{10}(x^3 - x^2) = \log_{5}(x^2) \]
This implies:
\[ x^3 - x^2 = x^2 \implies x^3 - 2x^2 = 0 \implies x^2(x - 2) = 0 \]
Since \(x > 1\), we have:
\[ x = 2 \]
Thus, \(x = y = z = 2\).

### Final Answer
The unique solution to the system is:
\[ x = y = z = 2 \]