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Find all third roots of 8^(1/3).

 Dec 22, 2015

Best Answer 

 #7
avatar+1904 
+10

8^(1/3)

 

(8+0i)^(1/3)

 

\(r=\sqrt(8^2+0^2)\)

 

\(r=\sqrt(64+0)\)

 

\(r=\sqrt(64)\)

 

\(r=8\)

 

\(tan(\Theta)=0/8\)

 

\(tan(\Theta)=0\)

 

\(\Theta=tan^-1(0)\)

 

\(\Theta=0\)

 

8e^(0i)^(1/3)

 

8^(1/3)e^(0i/3)

 

8^(1/3)*(cos(0)+i*sin(0))

 

8^(1/3)*(1+i*0)

 

8^(1/3)*(1+0)

 

8^(1/3)*1

 

8^(1/3)

 

2

 

8e^(2pi)^(1/3)

 

8^(1/3)e^(2pi/3)

 

8^(1/3)*(cos(2pi/3)+i*sin(2pi/3))

 

8^(1/3)*(-1/2+i*(sqrt(3)/2))

 

8^(1/3)*-1/2+i*sqrt(3)/2

 

2*-1/2+i*sqrt(3)/2

 

-2/2+i*sqrt(3)/2

 

-1+i*sqrt(3)/2

 

-1+sqrt(3)/2i

 

8e^(4pi)^(1/3)

 

8^(1/3)e^(4pi/3)

 

8^(1/3)*(cos(4pi/3)+i*sin(4pi/3))

 

8^(1/3)*(-1/2+i*(-sqrt(3)/2))

 

8^(1/3)*-1/2+i*-sqrt(3)/2

 

2*-1/2+i*-sqrt(3)/2

 

-2/2+i*-sqrt(3)/2

 

-1+i*-sqrt(3)/2

 

-1-sqrt(3)/2i

 Dec 22, 2015
 #1
avatar
+5

Find all third roots of 8^(1/3).

 

You only have ONE solution, which is 2.

 Dec 22, 2015
 #2
avatar
+5

The 1/3 root cancels out the 1/3 power and there can't be any other solutions because 1/3 roots only have one solution so the awnser is 8.

 Dec 22, 2015
edited by Guest  Dec 22, 2015
 #3
avatar+1904 
+5

There certainly are three roots.  The problem is 8^(1/3).

 Dec 22, 2015
 #4
avatar
+5

No. There are NOT. You are confusing third order variable such as x^3, with a cube root of a positive integer. Don't confuse the two.

 Dec 22, 2015
 #5
avatar
+5

The x in this equation has THREE solution, one real and two complex:      x^3-4x^2+6x-24 = 0

I think that is where you are going wrong.

 Dec 22, 2015
 #6
avatar+105606 
+10

Ther are 3 roots.

The first one is easy to find it is 2

 

There are 2pi radians in a circle and we want 3 roots so each will be 2pi/3 radians apart.

 

The roots will be

2,     2 cis(2pi/3)  and    2cis(-2pi/3)

 

 

 

\(cos(\frac{2\pi}{3}) = - cos (\frac{\pi}{3}) =\frac{ -1}{2}\\ sin(\frac{2\pi}{3})= sin(\frac{\pi}{3}) = \frac{\sqrt3}{2}\\ cis(\frac{2\pi}{3}) =\frac{ -1}{2}+i*\frac{\sqrt3}{2}\\ cis(\frac{2\pi}{3}) =\frac{ -1+\sqrt{3}\;i}{2}\\~\\ cos(\frac{-2\pi}{3}) = - cos (\frac{\pi}{3}) =\frac{ -1}{2}\\ sin(\frac{-2\pi}{3})= -sin(\frac{\pi}{3}) = \frac{-\sqrt3}{2}\\ cis(\frac{-2\pi}{3}) =\frac{ -1}{2}-i*\frac{\sqrt3}{2}\\ cis(\frac{-2\pi}{3}) =\frac{ -1-\sqrt{3}\;i}{2}\\ \)

 

\(\mbox{So the 3 solutions are }\\~\\ \sqrt[3]{8}=2,\qquad and \quad\\ \sqrt[3]{8} = -1+\sqrt{3}\;i\quad and \quad\\ \sqrt[3]{8} = -1-\sqrt{3}\;i, \\\)

 

.
 Dec 22, 2015
edited by Melody  Dec 22, 2015
edited by Melody  Dec 22, 2015
 #7
avatar+1904 
+10
Best Answer

8^(1/3)

 

(8+0i)^(1/3)

 

\(r=\sqrt(8^2+0^2)\)

 

\(r=\sqrt(64+0)\)

 

\(r=\sqrt(64)\)

 

\(r=8\)

 

\(tan(\Theta)=0/8\)

 

\(tan(\Theta)=0\)

 

\(\Theta=tan^-1(0)\)

 

\(\Theta=0\)

 

8e^(0i)^(1/3)

 

8^(1/3)e^(0i/3)

 

8^(1/3)*(cos(0)+i*sin(0))

 

8^(1/3)*(1+i*0)

 

8^(1/3)*(1+0)

 

8^(1/3)*1

 

8^(1/3)

 

2

 

8e^(2pi)^(1/3)

 

8^(1/3)e^(2pi/3)

 

8^(1/3)*(cos(2pi/3)+i*sin(2pi/3))

 

8^(1/3)*(-1/2+i*(sqrt(3)/2))

 

8^(1/3)*-1/2+i*sqrt(3)/2

 

2*-1/2+i*sqrt(3)/2

 

-2/2+i*sqrt(3)/2

 

-1+i*sqrt(3)/2

 

-1+sqrt(3)/2i

 

8e^(4pi)^(1/3)

 

8^(1/3)e^(4pi/3)

 

8^(1/3)*(cos(4pi/3)+i*sin(4pi/3))

 

8^(1/3)*(-1/2+i*(-sqrt(3)/2))

 

8^(1/3)*-1/2+i*-sqrt(3)/2

 

2*-1/2+i*-sqrt(3)/2

 

-2/2+i*-sqrt(3)/2

 

-1+i*-sqrt(3)/2

 

-1-sqrt(3)/2i

gibsonj338 Dec 22, 2015
 #8
avatar+104793 
+5

Thanks, Melody and gibsonj338   for those good answers.......this is a good one to learn from.......

 

 

 

cool cool cool

 Dec 22, 2015
 #9
avatar+105606 
+5

Guest, you are certainly correct. !

 

There is only 1 real roots and since most students only learn about real numbers you are 100% correct.

 

BUT

If you study maths for long enough you will learn about  imaginary. numbers.

 

Two of these roots are complex roots and that means that they are partly imaginary.

 

It is all based on the idea that    \(\sqrt{-1}=i\)

 

That is the basic element of the whole imaginary number system.

 

Lets see how this works,

 

I have said that one of the cubic roots is     \(-1-\sqrt3\;i\)

so if this is true then

\(2^3=(-1-\sqrt3\;i)^3\)


\(LHS=8\\ RHS=\;  (-1-\sqrt{3}\;i)(-1-\sqrt{3}\;i)(-1-\sqrt{3}\;i)\\ RHS=\; - \;\;\;\;(1+\sqrt{3}\;i)(1+\sqrt{3}\;i)\;\;\;\;(1+\sqrt{3}\;i)\\ RHS=\; -\;\;\;\; (1+2\sqrt3\;i+3*-1)\;\;\;\;(1+\sqrt{3}\;i)\\ RHS=\; -\;\;\;\; (1+2\sqrt3\;i-3)\;\;\;\;(1+\sqrt{3}\;i)\\ RHS= \;-\;\;\;\; (-2+2\sqrt3\;i)\;\;\;\;(1+\sqrt{3}\;i)\\ RHS= \;-\;\;\;\;-2 (1-\sqrt3\;i)\;\;\;\;(1+\sqrt{3}\;i)\\ RHS= \;\;\;\;\;2\;\; (1-\sqrt3\;i)\;\;\;\;(1+\sqrt{3}\;i)\\ RHS= \;\;\;\;\;2\;\; (1^2-(\sqrt3)^2(i^2))\\ RHS= \;\;\;\;\;2\;\; (1^2-(3)(-1))\\ RHS= \;\;\;\;\;2\;\; (1+3)\\ RHS=\;\;\;\;\;\ 8\\ RHS=LHS\\ \mbox{therefore one cubic root of 8 is }\:\: (-1-\sqrt{3}\;i) \)

.
 Dec 22, 2015
 #10
avatar+105606 
0

Thanks Chris :)

 Dec 22, 2015
 #11
avatar
+5

Being pedantic, you are all answering the wrong question.

You are answering the question 'What are the third roots of 8'.

The question actually asks for the third roots of 8^(1/3).

Assuming that all three values of 8^(1/3) are intended, it would seem that there are nine possibles.

 Dec 22, 2015
 #12
avatar+105606 
0

Yep you are right.

We often answer the question that we believe is intended rather than the one that is asked.

In this instance I did it without even realising that I was doing it : )

 

I am still pleased with my answers  wink

 Dec 22, 2015

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