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If we try to find the roots of $x^3-27,$ and we come up with $x=\sqrt[3]{27}=3,$ it is possible to contemplate the idea of having a root $(3)$ with mutiplicity $3.$ After all, the fundamental theorem of algebra calls for a maximum of 3 real roots - the alternative being 1 real and 2 complex.

Clearly, $(x-3)^3 \neq x^3-27,$ and the other two roots are complex: $-3/2i(\sqrt 3 - i)$ and its complex conjugate, $3/2 i (\sqrt 3 + i).$

Probably, if instead of just looking into $\sqrt[3]{27},$ I had gone directly to the cubic formula, I would have ended up with negatives under the roots. But the cubic formula is scary.

So the question is, how could I have guessed that instead of 1 real root with multiplicity of 3, I was going to end up with 1 real root and 2 complex roots?

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    $\begingroup$ Once you have the real root, divide by the respective linear factor and look at what's left. In this case it's really simple since $x^3-3^3=(x-3)(x^2+3x+3^2)\,$. $\endgroup$ – dxiv Jan 10 '18 at 23:34
  • $\begingroup$ @dxiv So, inspection is not enough... You have to divide, and then solve the roots of the resultant polynomial of degree $n-1?$ $\endgroup$ – Antoni Parellada Jan 10 '18 at 23:37
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    $\begingroup$ @user520393 That degree $n-1$ is $2$ in this case so, yes, just solve the quadratic. In the general case, you could also use Sturm's theorem to determine the number (and location) of real roots.. $\endgroup$ – dxiv Jan 10 '18 at 23:40
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    $\begingroup$ In case you are just talking about $x^3-27$: any polynomial of degree 3 with a root of multiplicity of 3 has the form $(x-a)^3=x^3+3ax^3+3a^2x+a^3$. When the quadratic and linear term are missing, but the constant term is not (i.e. $a\neq 0$), then you CAN'T have a multiplicity of 3. $\endgroup$ – SK19 Jan 10 '18 at 23:47
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    $\begingroup$ Notice (see Pedro's answer) that $e^{\frac{2\pi}{n}i}= -\frac{1}{2}+\frac{\sqrt 3}{2}i$ corresponds to the third root of unity $(n=3).$ These are $1$ when raised to the $n=3$ power, leaving the $r_1=3$ unaltered. The two additional roots are simply $3 \times e^{2\,\frac{2\pi}{n}i}$ and $3\times e^{\frac{2\pi}{n}i}.$ $\endgroup$ – Antoni Parellada Jan 11 '18 at 0:46
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In general you can rely on the following to simplify your life a bit:

Every irreducible polynomial over a perfect field (e.g. field of characteristic 0 and finite fields) is separable, which means that it has no multiple roots (all roots are different).

Therefore you have a very nice way to check wether it has multiple roots or not: factorize it in a convenient field, for example in $\mathbb{Q}$ where you can use criterions such as Eisenstein and so on. And once you have the factorization you should check that the factors do not have roots in common. This reduces at least the problem to a simpler one.

But in some cases you can already guess from the looks of the polynomial, as you said. In this case for example:

If the polynomial has the form $ x^{N}-a $ for some $a>0$ in $\mathbb{R}$ and $N>0$ in $\mathbb{N}$, then it has no multiple roots.

The reason: if $a>0$ then $\sqrt[N]{a}>0$ is a root already in $\mathbb{R}$. But now let $\zeta^{k}\in \mathbb{C}$ be the $N$ different $N^{th}$-roots of unity in $\mathbb{C}$. You know they are different because you can write them down explicitly with the exponential form of complex numbers for example, and you know those are all there are because of the fundamental theorem of algebra. Then, using that the powers of the product is the product of the powers and again arguing with the exponential form of complex numbers and the fundamental theorem of algebra, you see that the $N$ different roots of $x^{N}-a$ are $$ \zeta^{k}\sqrt[N]{a}$$ for $k=0,1,\ldots,N-1$.

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