which expression is a perfect cube?\n$x^{8}$\n$y^{24}$\n$m^{28}$\n$s^{64}$

Explanation:

Step1: Recall the rule for perfect - cube exponents

A variable with an exponent (n) is a perfect cube if (n) is divisible by 3.

Step2: Check the exponent of (x^8)

Since (8\div3=\frac{8}{3}), 8 is not divisible by 3, so (x^8) is not a perfect cube.

Step3: Check the exponent of (y^{24})

Since (24\div3 = 8), 24 is divisible by 3. So (y^{24}=(y^8)^3), and (y^{24}) is a perfect cube.

Step4: Check the exponent of (m^{28})

Since (28\div3=\frac{28}{3}), 28 is not divisible by 3, so (m^{28}) is not a perfect cube.

Step5: Check the exponent of (s^{64})

Since (64\div3=\frac{64}{3}), 64 is not divisible by 3, so (s^{64}) is not a perfect cube.

Answer:

(y^{24})