There are six different sixth roots of 64. That is, there are six complex numbers that solve \[x^6=64.\]Find them.

Guest Feb 3, 2018

#1**+1 **

Solve for x:

x^6 = 64

Take 6^th roots of both sides.

Taking 6^th roots gives 2 times the 6^th roots of unity:

**x = -2 or x = 2 or x = -2 (-1)^(1/3) or x = 2 (-1)^(1/3) or x = -2 (-1)^(2/3) or x = 2 (-1)^(2/3)**

Guest Feb 3, 2018

edited by
Guest
Feb 3, 2018

#2**+1 **

x^6 = 64

64 = 64 + 0i

The sixth root divides the unit circle into six even parts and the beginning angle is

arctan (0 / 64) = 0°

We have

^{6}√64 [ cos (0) + i sin (0)] = 2 [ 1 + 0 i ] = 2 + 0i

^{6}√64 [ cos (pi/ 3) + i sin (pi /3) = 2 [ 1/2 + √3/2 i ] = 1 + √3 i

^{6}√64 [ cos (2pi/ 3) + i sin (2pi/ 3) ] = 2 [ -1/2 + √3 i ] = - 1 + √3 i

^{6} √64 [ cos (pi) + i sin (pi) ] = 2 [ - 1 + 0 i ] = 2[-1] = -2 + 0i

^{6}√64 [ cos (4pi/3 + i sin (4pi/3) ] = 2 [ -1/2 - √3 / 2 i ] = - 1 - √3 i

6√64 [ cos (5 pi / 3) + i sin (5pi/ 3) ] = 2 [ 1/2 - i √3 /2 ] = 1 - √3 i

CPhill
Feb 3, 2018

#3**+1 **

It is also possible to use knowledge of difference and sum of squares and cubes to find all the possible solutions.

\(x^6=64\) | Move 64 to the other side. | |||||||

\(x^6-64=0\) | This can be written as a difference of squares. | |||||||

\(\left(x^3\right)^2-8^2=0\) | Since this is a difference of squares, factor it as such. | |||||||

\((x^3+8)(x^3-8)=0\) | Notice how both binomials are now difference and sum of cubes. Factor both completely. | |||||||

\((x+2)(x^2-2x+4)(x-2)(x^2+2x+4)=0\) | Now, set each factor equal to zero and solve. Let's do the easy ones first. | |||||||

| Now, let's tackle the harder ones. | |||||||

| Neither trinomial can be factored, so we will have to rely on other methods. I'll use the quadratic formula in this case. | |||||||

| Simplify both completely. | |||||||

| Now, convert the square root of a negative number to the simplest form in terms in i. | |||||||

| In both cases, the numerator and denominator both have common factors of two. | |||||||

| Since the original equation is written in the sixth degree, I know that the maximum number of roots equals the degree, or 6. This means I have found every root. | |||||||

Let's consolidate those roots together.

\(x_{1,2}=\pm 2\\ x_{3,4}=1\pm i\sqrt{3}\\ x_{5,6}=-1\pm i\sqrt{3}\)

TheXSquaredFactor
Feb 3, 2018

#4

#5**0 **

I do not see how this question is relevant to quotients, but I will evaluate the expression assuming that the "x" represents a multiplication operator.

6*4*60*68*45+45*23+34*34*5*67*78*90*98=266423797035

TheXSquaredFactor
Feb 6, 2018