If \(z=\sqrt[3]{9}\), what type of number is \(z^3\)?
Answer and explanation
Correct answer: Rational number
Since \(z=\sqrt[3]{9}\), we have \(z^3=(\sqrt[3]{9})^3=9\). The number 9 is an integer and therefore rational (9 = 9/1). Thus \(z^3\) is rational. Option B is wrong because 9 is not irrational; C is wrong because 9 is a real number (not non‑real); D is wrong because 9 is algebraic (it satisfies \(x-9=0\)), so it is not transcendental. Exam tip: simplify powers and roots first — often the result is an obvious integer or rational number after simplification.
Frequently asked questions
What is the correct answer to this question?
Rational number
Why is this the correct answer?
Since \(z=\sqrt[3]{9}\), we have \(z^3=(\sqrt[3]{9})^3=9\). The number 9 is an integer and therefore rational (9 = 9/1). Thus \(z^3\) is rational. Option B is wrong because 9 is not irrational; C is wrong because 9 is a real number (not non‑real); D is wrong because 9 is algebraic (it satisfies \(x-9=0\)), so it is not transcendental. Exam tip: simplify powers and roots first — often the result is an obvious integer or rational number after simplification.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Polynomials. Topic: Irrational numbers and real numbers.
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