To solve the question, we need to analyze the expression \(x = n(n+1)(n+2)\) where \(n\) is an even natural number, and check the divisibility of \(x\) and \(x^2\) by 48 and 144 respectively.

Step 1: Understanding the Expression

Since \(n\) is even, let \(n = 2k\) where \(k\) is an integer. Thus:

\(x = n(n + 1)(n + 2) = 2k(2k + 1)(2k + 2)\)

Rewriting \((2k + 2)\) as \(2(k + 1)\), we have:

\(x = 2k(2k + 1) \cdot 2(k + 1) = 4k(k+1)(2k+1)\)

Step 2: Determining Divisibility by 48

Divisibility by 48 requires the expression to have factors of 2⁴ and 3.

• We have at least four factors of 2 (from \(4k(k+1)\)), satisfying the power of 2 requirement.
• However, \(k\), \(k+1\), and \(2k+1\) do not necessarily provide a factor of 3. For a concrete example, if \(k = 2\), then \(2k + 1 = 5\) which is not divisible by 3.

Thus, Statement I is incorrect: \(x\) is not always divisible by 48.

Step 3: Determining Divisibility by 144

Divisibility by 144 requires \(x^2\) to be divisible by 144, which means:

\(x^2 = (4k(k+1)(2k+1))^2\) should be divisible by 144 = 2⁴ × 3².

• We already have \(x\) divisible by 16, hence \(x^2\) is divisible by 16² = 256, providing more than sufficient factors of 2.
• For divisibility by 9, at least one of \(k\), \(k+1\), or \(2k+1\) will be divisible by 3, ensuring that \(x^2\) will be divisible by 9.

Therefore, Statement II is correct: \(x^2\) is always divisible by 144.

Conclusion

The correct answer is:

II only

Thus, the statement that \(x^2\) is always divisible by 144 is correct, while the statement regarding divisibility of \(x\) by 48 is incorrect.