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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the cube root of 125 is 5, the cube root of 64 is 4, the cube root of 27 is 3, the cube root of 8 is 2, and the cube root of 1 is 1.
How about 27 whose cube root is 3 which is a rational whole number.
It means the number has three equal factors. 3 x 3 x 3 = 27 3 is the cube root of 27.
Cube root of 27 is 3 Cube root of 30 will be slightly higher than 3 (since 30 is higher than 27 by a just a little) which happens to be 3.11 (to 3 significant figure, 3.107... on a calculator)