Find all roots of $2x^3+16$ in $\mathbb C$ 
Find all roots of $p(x) = 2 x^3 + 16$ in $\mathbb C$.

I found my answers to be x = 2, -1+i$\sqrt{3}$, -1+i$\sqrt{3}$.
But when I put the expression into Symbolab, it gives me -2, 1+i$\sqrt{3}$, 1+i$\sqrt{3}$ as roots of p(x) in C.
Can someone explain where I went wrong?
This is how I did it;
First, I rewrite p(x) and get
$2x^3+16=0$
$2x^3=-16$
$x^3=-8$ ------ (1)
Express (1) in Euler form
$x=re^{\theta i}$
$x^3=r^3e^{3\theta i}$
Let w = -8
$ |w| = \sqrt{(-8)^2} = 8$
$\theta = tan^{-1}(0) = 0$
$w = 8e^{0\theta}$
Now I have 
$r^3e^{3\theta i} = 8e^{0\theta}$
Equate the modulus and argument
$r^3 = 8$
$r = 2$
$3\theta = 0+2\pi k$ ------- for k $\in Z$
$\theta = {2\pi k \over 3}$
Now I have
$ x = 2(cos {2\pi k \over 3} + i sin {2\pi k \over 3})$
Calculating x by substituting k = 0,1,2
k = 0, x = 2
k = 1, x = 2(${-1 \over 2} + i {\sqrt{3} \over 2}$) = -1+i${\sqrt{3}}$
k = 2, x = 2(${-1 \over 2} - i {\sqrt{3} \over 2}$) = -1-i${\sqrt{3}}$
So all roots of p(x) in C are 2, -1-i${\sqrt{3}}$, -1+i${\sqrt{3}}$
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I will be very appreciated if someone can help me out.

Another way you can calculate this is to use De Moivre's theorem, which states that if $z = r(\cos \theta + i\sin \theta)$, the $n$th roots of $z$ are $$r^{1/n}\left(\cos\frac{\theta+2\pi k}{n} + i \sin \frac{\theta+2\pi k}{n}\right)$$
From $(1)$, which is $x^3=-8$, we find that $z = 8(\cos \pi + i \sin \pi)$, so $r=8$, $\theta=\pi$ (because $8$ and $-8$ are opposite to each other), and $n=3$ (we are finding the $3$rd roots of $z$). Substituting these values into the formula, we have:
First root: $8^{1/3}(\cos \frac{\pi+2\pi *0}{3} + i \sin \frac{\pi+2\pi *0}{n}) = 2(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}) = 1+i\sqrt{3}$
Second root: $8^{1/3}(\cos \frac{\pi+2\pi *1}{3} + i \sin \frac{\pi+2\pi *1}{3}) = 2(\cos\pi+i \sin \pi) = -2$
Third root: $8^{1/3}(\cos \frac{\pi+2\pi *2}{3} + i \sin \frac{\pi+2\pi *2}{3}) = 2(\cos \frac{5\pi}{3} + i\sin\frac{5\pi}{3}) = 1-i\sqrt{3}$
Use the fact that $n$th roots of $z$ have $n$ complex roots. Have we have found all the complex $3$rd roots of $8$ yet?
