4
-21 + 1.7320508i1 - 1.7320508i
Some calculators don't have cube roots, so you could use X to the power of 1/root. x^(1/root). The cube root of 64 equals 4 because 4*4*4 (or 4^3) gives 64. You can reverse these powers by reciprocating the exponent: 64^(1/3) = 4
All numbers have cube roots (not necessarily integral cube roots) so every prime has cube roots.
They are 3 and 5 respectively
3√0.125 = 0.5 (there is only 1 real cube root of 0.125).
If the negative number is "-a", then you can say the cube root is "-(cube root of a)" Because if you cube a negative number, you get a negative number. So if you cube root a negative number, you get a negative number. Ex) cube root of -8 = -2 Because (-2)^3 = -8 But if you want to find the complex cube roots, you can make an equation: "x^3=-a" or "x^3+a=0" We know one of the roots is "-(cube root of a)" so you can factor the equation by (x+(cube root of a)) And then you use the quadratic formula for the quadratic equation you're left with. Ex) x^3=-8 or x^3+8=0 Since -2 is a root, factor it by (x+2) x^3+8=(x+2)(x^2-2x+4) Using the quadratic formula, you get "1+i√3" and "1-i√3" Therefore the three cube roots of -8 is <"-2", "1+i√3", "1-i√3">
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Square roots only have two solutions if the number is positive, if the number is negative it has no solutions.Actually ALL numbers (including negative numbers) have two square roots and three cube roots - just not all of them are real numbers; some are complex numbers.What is a complex number?A complex number is a number which is the sum of two parts: a real part and an imaginary part, ie are of the form (a + bi):a is the real part;bi is the imaginary part which is a real number (b) multiplied by the square root of -1 which is an imaginary value. To avoid having to write √-1 all the time, the little 'i' is used instead, ie i² = -1.Now the square roots of negative numbers can be found:eg √-4 = √(4 × -1) = √4 √-1 = 2i → the square root of -4 is ±2i.What about cube roots?³√8 = 2 is the real root, but there are two further complex roots: (-1 + √3 i) and (-1 - √3 i). I'll cube the first to show it does indeed equal 8 (remember that i² = -1)(-1 + √3 i)³ = (-1 + √3 i)(-1 + √3 i)(-1 + √3 i)= (-1 + √3 i)((-1)×(-1) + -1 × √3 i + -1 × √3 i + (√3 i)×(√3 i))= (-1 + √3 i)(1 - 2 (√3 i) + 3i²)= (-1 + √3 i)(1 - 2 (√3 i) + 3 × -1)= (-1 + √3 i)(1 - 2 (√3 i) - 3)= (-1 + √3 i)(-2 - 2 (√3 i))= (-1 + √3 i)(-1 - √3 i) × 2= ((-1)² - (√3 i)²) × 2= (1 - (-3)) × 2= (1 + 3) × 2= 4 × 2= 8Similar cubing of (-1 - √3 i) equals 8.
Because the cube of a positive number is positive and the cube of a negative number is negative.-------------------------------------------------------------------------------------------------------------------------------Every number has THREE cube roots. However, (at least) two of the three are complex numbers.For example, the cube roots of 8 are 2, (-1 + √3 i) and (-1 - √3 i) with i² = -1:2³ = 2 × 2 × 2 = 8(-1 + √3 i)³ = (-1 + √3 i)(-1 + √3 i)(-1 + √3 i)= (-1 + √3 i)((-1)² - 2√3 i + 3i²)= (-1 + √3 i)(1 - 2√3 i -3)= (-1 + √3 i)(-2 - 2√3 i)= (-1 + √3 i)(-1 - √3 i)2= ((-1)² - 3i²)2= (1 + 3)2= 4 × 2 = 8(-1 - √3 i)³ = (-1 - √3 i)(-1 - √3 i)(-1 - √3 i)= (-1 - √3 i)((-1)² + 2√3 i + 3i²)= (-1 - √3 i)(1 + 2√3 i -3)= (-1 - √3 i)(-2 + 2√3 i)= (-1 - √3 i)(-1 + √3 i)2= ((-1)² - 3i²)2= (1 + 3)2= 4 × 2 = 8
The ancients - Egyptians or Greeks. They probably came across the square root of 2 when considering the diagonal of a square with sides of length 1. The cube root of 3 would have arisen, similarly, with the principal diagonal of a unit cube.
They are 3 and 5 respectively
You do. The other two are complex numbers, of interest only tomathematicians and engineers, and usually not listed.For example, the three cube roots of 8 are:2-1 + i sqrt(3)-1 - i sqrt(3)
The 3 cube roots of a cubic equation are often going to require complex numbers, and this is one of them. The primary root is (-2) since (-2)3= -8 The other two roots are going to have the same mod (2) but be evenly spaced around the complex plane, so (2pi)/3 angular difference.These angles are pi/3 and 5pi/3(2,pi/3) = (1 + √3i) = (1 + 1.732i)(2,5pi/3) = (1 + -√3i) = (1- 1.732i)Check(-2)3 = -8◄(1 + √3i)3 = -8◄(1 - √3i)3 = -8◄■
The cube root of the expression 403 is 40 itself. This follows from the definition of cubes and cube roots.
The cube root for a real number is the value that when multiplied by itself three times produces that number. So that the cube root of B is A where A x A x A = B.Examplesthe cube root of 8 is 2... (2x2x2=8)the cube root of 27 is 3...(3x3x3=27)The special cases are 0 and 1, which are by definition their own roots (square roots, cube roots, etc.) because all powers of 0 and 1 are still the same value, zero or 1.Perfect cubes are numbers that can be cube-rooted into an integer.These include the numbers 0, 1, 8, 27, and so forth.Negative numbers can also be perfect cubes.-1x-1x-1=-1-2x-2x-2=-8-3x-3x-3=-27For non-perfect cubes, the root includes a repeating decimal requiring rounding.
there is no cube roots in negative