What is the cube root of 1728?
12
To find the cube root of 1728, express it as a product of prime factors: \(1728 = 2^6 \times 3^3\).
So, \(\sqrt[3]{1728} = \sqrt[3]{2^6 \times 3^3} = 2^2 \times 3 = 4 \times 3 = 12\).
Verification: \(12^3 = 12 \times 12 \times 12 = 1728\).
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