Factorize : $x^6 − 10x^3 + 27$ I want to factorize $$x^6 − 10x^3 + 27$$ 
I tried two methods , first I let $y=x^3 $ and converted it into a quadratic but the solutions are not real . 
The second method I tried was getting it to the form $a^3+b^3+c^3-3abc$ but I could only get close ( $(x^2)^3+(x)^3+3^3-3(x^2)(x)(3)$ but that is actually $x^6-8 x^3+27$ , close but not exact )  
I'm run out of ideas , so any help is appreciated .
Note : I know that the answer is $$(x^2+2 x+3) (x^4-2 x^3+x^2-6 x+9)$$ , but I want to know how !
 A: Here’s how I got the same factorization as @DietrichBurde:
I also thought of the polynomial as a polynomial in $x^3$, but I did not despair when I got the factorization $(x^3-(5+\sqrt{-2}\,)(x^3-(5-\sqrt{-2}\,)$, because I noticed that $5+\sqrt{-2}=(-1+\sqrt{-2}\,)^3$. (How did I notice this? I’ll reveal that at the end.)
Then each of the individual factors that I found by using the quadratic formula is a difference of cubes, so that, for instance the first of these has the factor $x-(-1+\sqrt{-2}\,)=x+1-\sqrt{-2}$. The other has a factor that is the conjugate of this, and the product of these two linear factors has integer coefficients, namely
$$
(x+1-\sqrt{-2}\,)(x+1-\sqrt{-2}\,)=x^2+2x+3\,,
$$
and this is necessarily a factor of our original polynomial. Now, the remaining quartic polynomial clearly has no further factorization, because if we call $-1+\sqrt{-2}=\alpha$ and a primitive cube root of unity $\omega$, then the roots of the quartic are certainly $\{\alpha\omega,\bar\alpha\omega,\alpha\bar\omega, \bar\alpha\bar\omega$, since $\omega\in\mathbb Q(\sqrt{-3}\,)$, a completely different field from $\mathbb Q(\sqrt{-2}\,)$. This means that the four quantities are a complete set of conjugates over $\mathbb Q$, in other words, no lower-degree $\mathbb Q$-polynomial can have any one of them as a root. This means that Dietrich’s factorization is the final story.
It only remains to know how I saw that $5+\sqrt{-2}$ was a cube in its integer ring. Here, I knew the arithmetic, namely that the ring has unique factorization, and when I saw that our number had norm $27$, I knew that it had to be the cube of something, a prime element, with norm $3$ (times a unit, but the only units are $\pm1$, and this could be absorbed into the cube root). Then it was just a matter of experimentation to see which of the numbers of norm $3$ cubed out to the desired number.
A: Seems to me in your attempted form $a^3+b^3 + c^3 - 3abc$ intuitively you came quite close, but quit too soon.
$x^6 − 10x^3 + 27 = (x^2)^3 + (2x)^3 +(3)^3 - 3\cdot (x^2) (2x)(3)  $
The rest I presume you can factorise.  
Of course such tricks work only on a few polynomials like the one in your question(s).
A: In such cases, we can start with 
$$(x^2)^3+(ax)^3+b^3-3\cdot x^2\cdot ax\cdot b=x^6+x^3(a^3-3ab)+b^3$$
Comparing with the given expression, $b^3=27, a^3-3ab=-10$
We know, the only real value of $b$ is $3\implies a^3-9a+10=0$
Clearly, $a=2$ satisfies the last equation 
A: Here is pen and paper solution of above question. @A Googler Just pay attention to $(x^2)^3+(ax)^3+b^3−3⋅x^2⋅ax⋅b=x^6+x^3(a^3−3ab)+b^3$
Instead of this for every question of MPA use only a no b and equate it to real equation.
All questions can be solved this way. Thanks to @lab bhattacharjee for giving the solution above. It's really elegant but no one noticed it.
A: Here is a backdoor way to get the quadratic factor.
Start with your factorization
$x^6-10x^3+27=(x^3-5+\sqrt{-2})(x^3-5-\sqrt{-2}).$
This is a factorization over $\mathbb Z[\sqrt{-2}]$, so seek a quadratic factor with the form
$x^2+2ax+(a^2+2b^2)=(x+a+b\sqrt{-2})(x+a-b\sqrt{-2}).$
The constant term in the quadratic polynomial has to divide $27$ and be positive; moreover $b=0$ cannot work since there are clearly no linear factors over the integers. Therefore $a^2+2b^2\in\{3,27\}$, from which $a=\pm1$ or $\pm5$ and $b=1$. So our candidates are limited to
$x^2+2x+3$
$x^2-2x+3$
$x^2+10x+27$
$x^2-10x+27.$
The third candidate fails because at $x=1$ it does not give divisors of $1^6-(10×1^3)+27=18$. The second and fourth candidates similarly fail at $x=2$. But the first candidate appears to give divisors of the sixth-degree polynomial for every absolutely small integer $x$, and when we try that factor by long division we find that it holds to give the quoted answer.
