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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
avatar
+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
avatar+87655 
+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

 

 

cool cool cool

CPhill  Feb 3, 2018
edited by CPhill  Feb 3, 2018
 #3
avatar+2143 
+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.
\(x+2=0\) \(x-2=0\)
\(x_1=-2\) \(x_2=2\)
   

 

  Now, let's tackle the harder ones.
\(x^2-2x+4=0\\ a=1\\ b=-2\\ c=4\) \(x^2+2x+4=0\\ a=1\\ b=2\\ c=4\)

 

  Neither trinomial can be factored, so we will have to rely on other methods. I'll use the quadratic formula in this case. 
\(x = {-(-2) \pm\sqrt{(-2)^2-4(1)(4)} \over 2(1)}\) \(x = {-(2) \pm\sqrt{(2)^2-4(1)(4)} \over 2(1)}\)

 

  Simplify both completely.
\(x = {2 \pm\sqrt{-12} \over 2}\) \(x = {-2 \pm\sqrt{-12} \over 2}\)

 

  Now, convert the square root of a negative number to the simplest form in terms in i.
\(x=\frac{2\pm 2i\sqrt{3}}{2}\) \(x=\frac{-2\pm 2i\sqrt{3}}{2}\)

 

  In both cases, the numerator and denominator both have common factors of two.
\(x_{3,4}=1\pm i\sqrt{3}\) \(x_{5,6}=-1\pm i\sqrt{3}\)

 

  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
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0

find the quotion.

6x4x60x68x45+45x23+34x34x5x67x78x90x98=

Guest Feb 5, 2018
 #5
avatar+2143 
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

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