The equation I'm trying to solve is $f(z) = 0$ where

$$f(z) = z^6 + z^3 + 1$$

I already tried the following: randomly throwing in complex numbers and real numbers, rational root theorem, banging my head on the table, and other painful things. Any ideas where I can start?

EDIT: Answer

Following Potato's hint, we have that if we set $y = z^3$ the resulting quadratic would be $y^2 + y + 1 = 0$, by which we can use the quadratic formula, so that

$$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

And then take the cube roots of the solutions. Thanks guys.

  • 10
    $\begingroup$ Set $y=z^3$, solve the resulting quadratic in $y$, the take cube roots of those solutions. $\endgroup$ – Potato Mar 11 '13 at 1:11
  • $\begingroup$ @Potato Thanks, I'll edit my question and put the answer in. $\endgroup$ – noobProgrammer Mar 11 '13 at 1:14

Hint: $(x^6+x^3+1)(x^3-1)=x^9-1$

| cite | improve this answer | |
  • $\begingroup$ Nice, Thomas Andrews! +1 $\endgroup$ – amWhy Mar 11 '13 at 1:16
  • $\begingroup$ +1 This is nice hint...subtle as an elephant in a crystal store, but very nice. $\endgroup$ – DonAntonio Mar 11 '13 at 5:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.