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What is the cube root of -8 with complex answers?
-21 + 1.7320508i1 - 1.7320508i
How do you do cube roots quickly?
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
What is prime with cube roots?
All numbers have cube roots (not necessarily integral cube roots) so every prime has cube roots.
What are the cube roots of 27 and 125?
They are 3 and 5 respectively
How do you find cube root of 1?
The real cube root of 1 is 1, since 13 = 1. There also a pair of complex cube roots.
What is the Real cube roots of 0.125?
3√0.125 = 0.5 (there is only 1 real cube root of 0.125).
What is the cube root of -8 with complex answers?
-21 + 1.7320508i1 - 1.7320508i
How do you do cube roots quickly?
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
What is prime with cube roots?
All numbers have cube roots (not necessarily integral cube roots) so every prime has cube roots.
How do you find the cube root of a negative number?
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">
How Many real cube roots dose 1 have?
1
Why do square roots have two solutions and cube roots only have one solution?
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.
Why does a cube root only have one value?
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
Who created square roots and cube roots?
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.
What are the cube roots of 27 and 125?
They are 3 and 5 respectively
Is 2 less than 9 raise to 1 over 3?
Yes.2^3 = 8 < 9 Taking cube roots, 2 < 9^(1/3)
How do you find cube root of 1?
The real cube root of 1 is 1, since 13 = 1. There also a pair of complex cube roots.
