How to solve $x^3=-1$? How to solve $x^3=-1$? I got following: 
$x^3=-1$ 
$x=(-1)^{\frac{1}{3}}$
$x=\frac{(-1)^{\frac{1}{2}}}{(-1)^{\frac{1}{6}}}=\frac{i}{(-1)^{\frac{1}{6}}}$...
 A: Set $\displaystyle x=re^{i \theta}$. So $\displaystyle r^3e^{i3\theta}= x^3= -1= e^{i \pi}$, hence $r^3=1$ and $3 \theta= \pi [2\pi]$. Finally, $r=1$ and $\displaystyle \theta = \frac{\pi}{3} \left[\frac{2\pi}{3} \right]$ so you get three solutions: $$x_1= e^{i \pi/3}= \frac{1}{2}+ i \frac{\sqrt{3}}{2}, \ x_2=e^{-i 2 \pi /3}= \frac{1}{2} - i \frac{\sqrt{3}}{2}, \ x_3= e^{i \pi}=-1.$$
A: $x^3+1=0\implies (x+1)(x^2-x+1)=0$
If $x+1=0,x=-1$
else $x^2-x+1=0$ then  $x=\frac{1±\sqrt{1-4}}{2\cdot1}=\frac{1±\sqrt3i}{2}$ using the well known Quadratic Formula. 
Alternatively, using Euler's identity  and Euler's formula like other two solutions,
  for $x^n=-1=e^{i(2m+1)\pi}$ as $e^{i\pi}=-1$ where $m$ any any integer, $n$ is a natural number .
We know $n$-degree equation has exactly $n$ roots, so the roots of $x^n+1=0$
are $e^{\frac{i(2m+1)\pi}{n}}=\cos\frac{(2m+1)\pi}{n}+i\sin \frac{(2m+1)\pi}{n}$ where  $m=0,1,2,...n-1$. It's just customary, not mandatory that we have defined $m$  to assume this range of values, in fact it can assume any $n$ in-congruent values (for example, consecutive values like $r,r+1,...,r+n-1,$  where r is any integer), the reason is explained below.
(1)Using Repeated Root Theorem, let $f(x)=x^n+R$ where $n>1$ and $R ≠ 0\implies x≠0$
So, $\frac{df}{dx}=nx^{n-1}, \frac{df}{dx}=0$ does not have any non-zero root.
Clearly, $f(x)=x^n+R$ can not have any repeated root unless $R=0$
(2)Let $e^{\frac{i(2a+1)\pi}{n}}=e^{\frac{i(2b+1)\pi}{n}}$
$\implies e^{\frac{2i(a-b)\pi}{n}}=1=e^{2k\pi i}$ where $k$ is any integer.
$\implies a-b=kn\implies a≡b\pmod n$, so any $n$ in-congruent values of $m$ will give us essentially the same set of n distinct solutions.
Here $n=3$ so, let's take $m=0,1,2$.
$m=0\implies  \cos\frac{\pi}{3}+i\sin \frac{\pi}{3}=\frac{1+i\sqrt3}{2} $
$m=1\implies  \cos \pi+i\sin\pi=-1 $
$m=2\implies  \cos\frac{5\pi}{3}+i\sin \frac{5\pi}{3}=\frac{1-i\sqrt3}{2} $
A: Observe that $(e^{a i})^3 = e^{3 a i}$ and $-1=e^{\pi i}$.
A: $$x^3=-1$$
$$x^3+1=0$$
$$(x+1)(x^2+1-x)=0$$
$$x=-1 \quad\text{or}\quad x^2-x+1=0$$
When $$x^2-x+1=0, x= \frac{-1(\pm\sqrt{-3})}{2}$$
$$x= \frac{-1+\sqrt{3}i}{2} \quad\text{and}\quad \frac{-1-\sqrt{3}i}{2}$$
Which is equal to $e^{−i2π/3}$ and $e^{i2π/3}$ respectively. 
A: Let $x=a+bi$, where $a$ and $b$ are real.  Then $(a+bi)^3 = -1$.  Expanding the left-hand side gives $$a^3 +3a^2bi -3ab^2 -b^3i = -1.$$
We can separate the real and imaginary parts of this equation:
$$\begin{align}
a^3 -3ab^2 & = -1 \\
3a^2b - b^3 & = 0
\end{align}$$
Taking the obvious $b=0$ solution gives us $a=-1$ and thus the real solution

$$x =-1.$$

So suppose $b\ne 0$. Then the second equation has $3a^2 = b^2$, so $b = \pm a\sqrt3$. Putting $3a^2$ for $b^2$ in the first equation gives $$ a^3 - 9a^3 = -1$$ so $a = \frac12.$ Since $b = \pm a\sqrt3$, we have $b=\pm\frac{\sqrt3}2$. This gives us the other two solutions, namely 

$$x=\frac12 \pm\!\frac{\sqrt3}2i.$$

A: The real solution:
$$x^3+1=0<=>$$
$$x^3=-1<=>$$
Take cube roots of both sides:
$$x=-1$$
Complex solution:
$$x^3+1=0<=>$$
$$x^3=-1<=>$$
$$x^3=|-1|e^{arg(-1)i}<=>$$
$$x^3=e^{(\pi +2\pi k)i}<=>$$
(with k is the element of Z)
$$x=\left(e^{(\pi +2\pi k)i}\right)^{\frac{1}{3}}<=>$$
$$x=e^{\left(\frac{1}{3}\pi +\frac{2}{3}\pi k\right)i}$$
(with k goes from 0 to 3 -> k=0-2)
A: Complex numbers have a great visual interpretation. $x+ iy$ is visualized as the point in the plane with coordinates $(x,y)$. When you multiply two complex numbers, you "add the angles and multiply the lengths". This visual interpretation of complex number multiplication allows us to find the solutions to $x^3 = -1$ immediately, with no writing. Of course $-1$ is one solution. You probably see another solution already. It's on the unit circle and it makes an angle of 60 degrees with the x-axis. And finally you see the third solution as well.
It's a fact that polynomials of degree $3$ have at most $3$ distinct roots, so we have found all the solutions to $x^3+1=0$.
A: $$x^3+1=0$$
$$(x+1)(x^2-x+1)=0$$
$$x=-1,\frac{1\pm \sqrt{3}i}{2}$$
$$x=e^{i\pi},e^{i2\pi/3},e^{-i2\pi/3}$$
