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I have two problems that I have been trying to find the roots of:$ x^6-8x^3$ and $x^6-1$. I don't have the first real clue on how to solve the first polynomial after taking out $x^3$. I tried using the rational zero theorem on the second, but I keep getting more polynomials I have to just use the theorem on again. Can any one help?

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  • $\begingroup$ You could also substitute $u=x^3$, solve the resulting quadratics, and then "unsubstitute". $\endgroup$ – Χpẘ Apr 27 '17 at 1:40
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For $x^6-8x^3$,$$x^3(x^3-8)=0$$ will suffice. where there will be 4 real and 2 imaginary roots

However, the equation $$x^6=1$$ is pointing towards $6$ roots of unity, giving you two real root and 4 imaginary roots: $$e^{i\frac{\pi}{6}},e^{i\frac{2\pi}{6}},e^{i\frac{3\pi}{6}},e^{i\frac{4\pi}{6}} \text{ and } e^{i\frac{5\pi}{6}}$$

Here $e^{i\frac{\pi}{2}}$ is real root

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    $\begingroup$ 6 roots of unity, giving you one real root and 5 imaginary roots No. More like $2+4$. $\endgroup$ – dxiv Apr 27 '17 at 1:34
  • $\begingroup$ Indeed, My bad @dxiv $\endgroup$ – The Dead Legend Apr 27 '17 at 1:35
  • $\begingroup$ @TheDeadLegend You have the same root $e^{i\pi/6}$ listed twice. $\endgroup$ – Χpẘ Apr 27 '17 at 1:36
  • $\begingroup$ Also, your list is not quite right: It omits $1$, and lists some roots of $x^{12} = 1$. You should have $e^0, e^{i\pi/3}, e^{2i\pi/3}, e^{i\pi}, e^{4i\pi/3}, e^{5i\pi/3}$. $\endgroup$ – Brian Tung Apr 27 '17 at 1:37
  • $\begingroup$ The real roots are $e^0 = 1$ and $e^{i\pi} = -1$. Your arguments are off by a factor of $2$. $\endgroup$ – Brian Tung Apr 27 '17 at 1:40
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Hints (using just elementary factorization and quadratics):

  • $x^6-8x^3 = x^3(x^3-8)=x^3(x-2)(x^2+2x+4)$

  • $x^6-1 = (x^3-1)(x^3+1)=(x-1)(x^2+x+1)(x+1)(x^2-x+1)$

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We have $f(x) = x^6-8x^3$ and $g(x) = x^6-1$

So $f(x) = x^3(x^3-8)$

Notice that $x^3=8$ is a root which implies $x=2$ is a root.

Then $f(x) = x^3(x-2)(x^2+ax+4)\quad$ where $a$ is a constant we should find.

Expanding this out we get $f(x) = x^3(x^3+(a-2)x^2+(4-2a)x-8)$

Equating coefficients we can see that $a-2=0\implies a=2$

So $f(x) = x^3(x-2)(x^2+2x+4) = x^3(x-2)((x+1)^2+3)$

Then using difference of two squares:

$f(x) = x^3(x-2)(x+1+\sqrt{3}i)(x+1-\sqrt{3}i)$

We can do the same trick with $g(x) = (x^3)^2-(1)^2$

$\implies g(x) = (x^3+1)(x^3-1)$

Then note that $x^3=\pm 1$ are roots which implies that $x=\pm 1$ are roots too.

$\implies g(x) = (x-1)(x^2+ax+1)(x+1)(x^2+bx+1)$ where $a$ and $b$ are constants we should find.

By comparing coefficients, we can find $-a+1=0\implies a=1$

and $b+1=0\implies b=-1$

Then $g(x) = (x-1)(x^2+x+1)(x+1)(x^2-x+1)$

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