Statement-1: `If a, b, c in Q and 2^(1//3)` is a root of `ax^(2) + bx + c = 0`, then a = b = c = 0.
Statement-2: A polynomial equation with rational coefficients cannot have irrational roots.

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Explanation

To solve the problem, we need to analyze both statements provided in the question.

**Step 1: Analyze Statement 1**
Statement 1 claims that if \( a, b, c \in \mathbb{Q} \) and \( 2^{1/3} \) is a root of the quadratic equation \( ax^2 + bx + c = 0 \), then \( a = b = c = 0 \).

1. **Assumption**: Let \( a, b, c \) be rational numbers (i.e., \( a, b, c \in \mathbb{Q} \)).
2. **Substituting the root**: Since \( 2^{1/3} \) is a root, substituting \( x = 2^{1/3} \) into the equation gives:
\[ a(2^{1/3})^2 + b(2^{1/3}) + c = 0 \]
Simplifying this, we have:
\[ a(2^{2/3}) + b(2^{1/3}) + c = 0 \]

3. **Multiplying by the LCM**: To eliminate the irrational terms, we can multiply the entire equation by \( 2^{3/3} \) (which is 2) to obtain:
\[ 2a(2^{2/3}) + 2b(2^{1/3}) + 2c = 0 \]
This simplifies to:
\[ 2a(2^{2/3}) + 2b(2^{1/3}) + 2c = 0 \]

4. **Analyzing the coefficients**: Since \( a, b, c \) are rational, the left-hand side must also be rational. However, \( 2^{1/3} \) and \( 2^{2/3} \) are irrational, leading to a contradiction unless \( a = b = c = 0 \).

5. **Conclusion for Statement 1**: The only solution that avoids contradiction is \( a = b = c = 0 \). Thus, Statement 1 is **true**.

**Step 2: Analyze Statement 2**
Statement 2 states that a polynomial equation with rational coefficients cannot have irrational roots.

1. **Counterexample**: Consider the polynomial \( x^3 - 2 = 0 \). This polynomial has rational coefficients (1 and -2).
2. **Finding the roots**: The root of this polynomial is \( x = 2^{1/3} \), which is irrational.
3. **Conclusion for Statement 2**: Since we found a polynomial with rational coefficients that has an irrational root, Statement 2 is **false**.

**Final Conclusion**:
- Statement 1 is **true**.
- Statement 2 is **false**.

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