Question 1 • OK-Algebra II A
A perfect cube has a numerical factor that is a cube and variable exponents that are multiples of 3. Since −1,000 = (−10)³, choice A is a perfect cube; 27r²¹ = (3r⁷)³, so B is one as well. Also, g³h³⁰ = (gh¹⁰)³, and −1 = (−1)³, making D and E perfect cubes. Choice C, 343p, is not a perfect cube because the exponent of p is 1, not a multiple of 3.
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