Is `{x:x^(3) +1 =0,x in N}=phi` true where N is the set of positive integers ?

Answer

To determine whether the set \( \{ x : x^3 + 1 = 0, x \in \mathbb{N} \} = \emptyset \) is true, we will follow these steps:

### Step 1: Understand the equation
We start with the equation given in the set:
\[ x^3 + 1 = 0 \]
This can be rewritten as:
\[ x^3 = -1 \]

### Step 2: Solve for \( x \)
To find \( x \), we take the cube root of both sides:
\[ x = \sqrt[3]{-1} \]
This gives us:
\[ x = -1 \]

### Step 3: Check if \( x \) belongs to \( \mathbb{N} \)
The problem states that \( x \) must belong to the set of natural numbers \( \mathbb{N} \), which is defined as the set of positive integers:
\[ \mathbb{N} = \{ 1, 2, 3, \ldots \} \]
Since \( -1 \) is not a positive integer, we conclude that there are no values of \( x \) in \( \mathbb{N} \) that satisfy the equation \( x^3 + 1 = 0 \).

### Step 4: Conclusion about the set
Since there are no elements \( x \) that satisfy the equation and belong to \( \mathbb{N} \), the set can be expressed as:
\[ \{ x : x^3 + 1 = 0, x \in \mathbb{N} \} = \emptyset \]

### Final Answer
Thus, we conclude that the statement \( \{ x : x^3 + 1 = 0, x \in \mathbb{N} \} = \emptyset \) is **true**.

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