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I would like to factorize $a^6+8a^3+27$.

I got different answers but one of the answers is

$$(a^2-a+3)(a^4+a^3-2a^2+3a+9)$$

Can someone tell me how to get this answer? Thanks.

Olivier Oloa
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Abhishek Kasargod
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    You can write $b=a^3$ so it becomes $b^2+8b+27$ which you can factor as a quadratic. Then substitute $a^3$ back in and see if you can factor either of those cubics. – Gregory Grant Jul 28 '16 at 15:09
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    @GregoryGrant, $b^2+8b+27=(b+4+\sqrt{-11})(b+4-\sqrt{-11})$. How does that help? – Barry Cipra Jul 28 '16 at 16:11
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    @BarryCipra it helps because you've now got two cubics instead of one sixth degree. In fact an easy cubic because it is of the form $a^3+c$, the roots are $\sqrt[3]{c}$ times the three possible roots of unity. So yes that gets you all the way there to six linear factors. – Gregory Grant Jul 28 '16 at 16:23
  • @GregoryGrant, I'd really like to see how you get from those six linear factors back to a factorization over the integers. – Barry Cipra Jul 28 '16 at 16:28
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    @BarryCipra Easy, two factors will be complex conjucates of each other, multiply those together, then multiply together the other four. That will give you alternate factorizations including the one above. – Gregory Grant Jul 28 '16 at 18:56
  • @GregoryGrant, I'd still like to see the details of this approach. After all, what you've described works just as easily for finding a factorization of $a^6+8a^3+26$.... – Barry Cipra Jul 28 '16 at 19:24

5 Answers5

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$$a^6+8a^3+27= (a^2)^3+3^3+(-a)^3-3\cdot a^2\cdot3(-a)$$

Now $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$

See : If $a,b,c \in R$ are distinct, then $-a^3-b^3-c^3+3abc \neq 0$.

Community
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lab bhattacharjee
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The roots of $a^6+8a^3+27$ are the cubic roots of the roots of $x^2+8x+27$,
given by $-4\pm\sqrt{-11}$. If we set $\alpha=-4+\sqrt{-11}=\left(\frac{1-\sqrt{-11}}{2}\right)^3$ and $\omega=\exp\left(\frac{2\pi i}{3}\right)$,
it follows that:

$$ a^6+8a^3+27 = \color{red}{\left(a-\alpha^{1/3}\right)\left(a-\bar{\alpha}^{1/3}\right)}\color{blue}{\left(a-\omega\alpha^{1/3}\right)\left(a-\omega\bar{\alpha}^{1/3}\right)\left(a-\omega^2\alpha^{1/3}\right)\left(a-\omega^2\bar{\alpha}^{1/3}\right)}$$ and we just have to understand how to "couple back" such factors (as suggested by colors),
that is easy through the identities $$ \alpha^{1/3}+\bar{\alpha}^{1/3}=1,\qquad \alpha\bar{\alpha}=3^3,\qquad \omega+\bar{\omega}=-1. $$

Jack D'Aurizio
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  • I must be having another off day, because I don't see how you got the first identity, nor how the identities help pick out which pair of complex conjugates give the quadratic factor $a^2-a+3$. – Barry Cipra Jul 28 '16 at 19:34
  • @BarryCipra: there was a miscomputation on my side, but I hope everything is easier to follow now. – Jack D'Aurizio Jul 28 '16 at 20:54
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    Ah, thanks! So the secret is recognizing $-4+\sqrt{-11}$ as a cube of something nice. – Barry Cipra Jul 28 '16 at 21:01
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One may write $$ \begin{align} a^6+8a^3+27&=a^3\left(a^3+\frac{27}{a^3}+8 \right) \\&=a^3\left(\left(a+\frac3a\right)^3-9\left(a+\frac3a\right)+9-1 \right) \\&=a^3\left(\left[\left(a+\frac3a\right)^3-1\right]-9\left[\left(a+\frac3a\right)-1\right] \right) \\&=a^3\left(a+\frac3a-1\right)\left(\left(a+\frac3a\right)^2+\left(a+\frac3a\right)-8 \right) \\&=a \cdot \left(a+\frac3a-1\right)\cdot a^2 \cdot\left(\left(a+\frac3a\right)^2+\left(a+\frac3a\right)-8 \right) \\&=\left(a^2-a+3\right)(a^4+a^3-2a^2+3a+9). \end{align} $$

Olivier Oloa
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Here's a slightly sneaky way to arrive at the factorization.

Let $P(a)=a^6+8a^3+27$. Then

$$P(10)=1008027=3\cdot3\cdot31\cdot3613=93\cdot3\cdot3613=(10^2-10+3)\cdot3\cdot3613$$

so we might guess that $a^2-a+3$ is a factor of $a^6+8a^3+27$. If it is, then the other factor would have to take the form $a^4+a^3+\cdots+9$ (so that the product with $a^2-a+3$ has no $a^5$ term and ends with a $27$), and we can see that

$$3\cdot3613=10839=10^4+10^3-2\cdot10^2+3\cdot10+9$$

suggests $a^4+a^3-2a^2+3a+9$ as the other factor.

Note, one might also try grouping $P(10)=9\cdot31\cdot3613=(10-1)\cdot31\cdot3613$, but $a-1$ is clearly not a factor. (Note also that the hard work that's been swept under the rug here is the work that goes into factoring $1008027$.)

Barry Cipra
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By the rational root theorem, we guess the factors include at least one of the following:

$$x \pm 27$$

$$x \pm 9$$

$$x \pm 3$$

$$x \pm 1$$

Where did I get those?

  1. Take the absolute value of the last term (in descending order), $27$, and the absolute value of the last term, $1$.

  2. Take the factors of each:

$$27: \color{red}{1,3,9,27}$$

$$1: \color{green}{1}$$

  1. Consider $x-a$ where $a$ may be any of the ff:

$$\pm \frac{\color{red}{27}}{\color{green}{1}}$$

$$\pm \frac{\color{red}{9}}{\color{green}{1}}$$

$$\pm \frac{\color{red}{3}}{\color{green}{1}}$$

$$\pm \frac{\color{red}{1}}{\color{green}{1}}$$

BCLC
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