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

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

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

You only have ONE solution, which is 2.

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.

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

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.

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.

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)

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

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

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

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

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

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

Thanks Chris :)

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.

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