Which of the following is true for (\sqrt[3]{16})?
Answer and explanation
Correct answer: It is irrational
Prime factorization gives 16 = 2^4. A number is a perfect cube only if every prime exponent is a multiple of 3. Since 4 is not divisible by 3, \(\sqrt[3]{16}\) is not a perfect cube and therefore is irrational. Options B and C are incorrect because 4^3 = 64 and 8^3 = 512, not 16. Option D is wrong because an irrational number cannot be expressed as a terminating decimal. Exam tip: to test whether \(\sqrt[k]{N}\) is rational, factor N and check if every prime exponent is divisible by k.
Frequently asked questions
What is the correct answer to this question?
It is irrational
Why is this the correct answer?
Prime factorization gives 16 = 2^4. A number is a perfect cube only if every prime exponent is a multiple of 3. Since 4 is not divisible by 3, \(\sqrt[3]{16}\) is not a perfect cube and therefore is irrational. Options B and C are incorrect because 4^3 = 64 and 8^3 = 512, not 16. Option D is wrong because an irrational number cannot be expressed as a terminating decimal. Exam tip: to test whether \(\sqrt[k]{N}\) is rational, factor N and check if every prime exponent is divisible by k.
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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