which monomial is a perfect cube?\n1x³\n3x³\n6x³\n9x³
which monomial is a perfect cube?\n1x³\n3x³\n6x³\n9x³
Answer
Explanation:
Step1: Recall perfect - cube rule
A monomial (ax^{n}) is a perfect cube if (a) is a perfect - cube number and (n) is divisible by 3. The cube of an integer (m) is (m^{3}), and for a monomial in the form (ax^{n}), we check the coefficient (a).
Step2: Check each coefficient
- For the coefficient of (1x^{3}), since (1 = 1\times1\times1=1^{3}), and the exponent of (x) is 3 which is divisible by 3.
- For the coefficient of (3x^{3}), (3) is not a perfect - cube number because there is no integer (m) such that (m^{3}=3).
- For the coefficient of (6x^{3}), (6) is not a perfect - cube number because there is no integer (m) such that (m^{3}=6).
- For the coefficient of (9x^{3}), (9) is not a perfect - cube number because there is no integer (m) such that (m^{3}=9).
Answer:
(1x^{3})