Let `a, n in N` such that `a ge n^(3)`. Then `root3(a+1)-root3(a)` is always

Options

A. less than `(1)/(3n^(2))`
B. less than `(1)/(2n^(3))`
C. more than `(1)/(n^(3))`
D. more than `(1)/(4n^(2))`

Explanation

To solve the problem, we need to analyze the expression \( \sqrt[3]{a+1} - \sqrt[3]{a} \) under the condition that \( a \geq n^3 \) for some natural number \( n \).

### Step 1: Define the function
Let \( f(x) = \sqrt[3]{x} \). We want to evaluate \( f(a+1) - f(a) \).

### Step 2: Apply the Mean Value Theorem
According to the Mean Value Theorem, if \( f \) is continuous on the closed interval \([a, a+1]\) and differentiable on the open interval \((a, a+1)\), then there exists some \( c \in (a, a+1) \) such that:
\[ f'(c) = \frac{f(a+1) - f(a)}{(a+1) - a} = f(a+1) - f(a) \]

### Step 3: Differentiate the function
Now, we need to find the derivative of \( f(x) \):
\[ f'(x) = \frac{1}{3} x^{-\frac{2}{3}} = \frac{1}{3 \sqrt[3]{x^2}} \]

### Step 4: Evaluate the derivative at \( c \)
From the Mean Value Theorem, we have:
\[ \sqrt[3]{a+1} - \sqrt[3]{a} = f'(c) = \frac{1}{3 \sqrt[3]{c^2}} \]
where \( c \) is in the interval \( (a, a+1) \).

### Step 5: Analyze the value of \( c \)
Since \( a \geq n^3 \), it follows that \( c \) will also be greater than \( n^3 \) because \( c \) lies between \( a \) and \( a+1 \). Therefore, we can conclude:
\[ c \geq n^3 \]

### Step 6: Substitute \( c \) into the derivative
Now substituting \( c \) into the expression for \( f'(c) \):
\[ \sqrt[3]{a+1} - \sqrt[3]{a} = \frac{1}{3 \sqrt[3]{c^2}} \leq \frac{1}{3 \sqrt[3]{(n^3)^2}} = \frac{1}{3 \sqrt[3]{n^6}} = \frac{1}{3 n^2} \]

### Conclusion
Thus, we have shown that:
\[ \sqrt[3]{a+1} - \sqrt[3]{a} \leq \frac{1}{3 n^2} \]
This means that \( \sqrt[3]{a+1} - \sqrt[3]{a} \) is always less than or equal to \( \frac{1}{3 n^2} \).