`y=(x^(3)+(x^(6))/(2)+(x^(9))/(3)+……)` then

Options

A. `x^(3)=1-e^(-y)`
B. `x=log(x+y)`
C. `x^(3)=e^(y)`
D. `x=1+e^(y)`

Answer

To solve the given problem, we need to find the value of \( y \) defined by the series:

\[ y = x^3 + \frac{x^6}{2} + \frac{x^9}{3} + \ldots \]

### Step 1: Identify the series
The series can be rewritten in a more general form. Notice that the terms are of the form:

\[ y = \sum_{n=1}^{\infty} \frac{x^{3n}}{n} \]

### Step 2: Recognize the series as a logarithmic series
The series \(\sum_{n=1}^{\infty} \frac{x^n}{n}\) is known to converge to \(-\log(1-x)\) for \(|x| < 1\). Therefore, we can relate our series to this known result.

### Step 3: Substitute \(x^3\) into the logarithmic series
By substituting \(x^3\) into the logarithmic series, we have:

\[ y = \sum_{n=1}^{\infty} \frac{(x^3)^n}{n} = -\log(1 - x^3) \]

### Step 4: Solve for \(y\)
Thus, we can express \(y\) as:

\[ y = -\log(1 - x^3) \]

### Step 5: Exponentiate both sides
To eliminate the logarithm, we exponentiate both sides:

\[ e^{-y} = 1 - x^3 \]

### Step 6: Rearranging the equation
Rearranging gives us:

\[ x^3 = 1 - e^{-y} \]

### Conclusion
This means that the relationship we derived is:

\[ x^3 + e^{-y} = 1 \]

### Final Answer
Thus, the final expression for \(y\) in terms of \(x\) is:

\[ e^{-y} = 1 - x^3 \]