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The cube root of 27 is 3. 27 has three cube roots, a real cube root (3) and two imaginary cube roots. To compute them, start with the following equation and solve for x: x^3 = 27 x^3 - 27 = 0 Factor using the difference of cubes: (x - 3)(x² + 3x + 9) = 0 One root is x = 3. The other two can be found using the quadratic formula on: x² + 3x + 9 = 0 a = 1 b = 3 c = 9 x = [-3 ± √(9 - 4*1*9) ] / 2 x = (-3 ± √(-27) ] / 2 x = (-3 ± 3i√3 ) / 2 x = -3/2 ± (3i√3)/2 Answers: 3 -3/2 + (3i√3)/2 -3/2 - (3i√3)/2 Hope I helped :)

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13y ago

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