which monomial is a perfect cube?\n16x^6\n27x^8\n32x^12\n64x^6

Explanation:

Step1: Recall perfect - cube rules

A number (a) is a perfect cube if (a = b^3) for some integer (b), and for a variable (x^n), it is a perfect cube if (n) is divisible by 3.

Step2: Analyze each option

Option 1: (16x^{6})

The number 16 is not a perfect cube since (16=2^4), and (\sqrt[3]{16}) is not an integer.

Option 2: (27x^{8})

The number 27 is a perfect cube ((27 = 3^3)), but the exponent of (x) is 8, and (8\div3=\frac{8}{3}), so (x^8) is not a perfect - cube power of (x).

Option 3: (32x^{12})

The number 32 is not a perfect cube since (32 = 2^5), and (\sqrt[3]{32}) is not an integer.

Option 4: (64x^{6})

The number 64 is a perfect cube ((64=4^3 = 2^6)), and the exponent of (x) is 6, and (6\div3 = 2), so (x^6=(x^2)^3).

Answer:

(64x^{6})