Solve for $x$, $$27x^3+21x+8=0$$

I would like to know if there exists an formula for cubic equations just like quadratic formula for quadratic equations.

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    $\begingroup$ Cardano's formula works, but it is not exactly easy to use. Your example has a rational root, so it is best to start there. $\endgroup$ – lulu Feb 16 at 14:19
  • $\begingroup$ Not clear what you want by way of more details, Robert. Could you be more explicit about what is missing in the answers that have been posted? $\endgroup$ – Gerry Myerson Feb 19 at 10:09
  • $\begingroup$ Please, Robert, what more do you need? $\endgroup$ – Gerry Myerson Feb 21 at 0:33
  • $\begingroup$ Sorry for delay in replying $\endgroup$ – user751264 Feb 21 at 16:56
  • $\begingroup$ I am actually looking for a method which can be always applied (as the quadratic formula)to solve a cubic almost instantaneously $\endgroup$ – user751264 Feb 21 at 16:58

We can begin by making an general cubic equation into a depressed cubic( missing square term)

To make $$ax^3+bx^2+cx+d=0$$ a depressed cubic we substitute $$x=y-\frac b{3a}$$ to get $$a {\left[y-\frac b{3a}\right]}^3+b{\left[y-\frac b{3a}\right]}^2+c{\left[y-\frac b{3a}\right]}+d=0$$

Which simplyfies to $$ay^3+y \left[c-\frac b{3a}\right]+\left[d+\frac{2b^3}{27a^2}-\frac{bc}{3a}\right]=0$$

Now we have a depressed cubic so now we can just simply use the shortened Cardano's formula which is, $$ x={\left[\frac{-d}2+\sqrt{\frac{d^2}4+ \frac{c^3}{27}}\right]}^\frac13 +{\left[\frac{-d}2-\sqrt{\frac{d^2}4+ \frac{c^3}{27}}\right]}^\frac13 $$

Just put up the values and answer will pop out!

P.S. It took a hell lot of formatting to do this. Anyone knows how to do it faster?.


Better way to do it is this. If you write $t=3x$ you get $$t^3+7t+8=0$$

so $$t^3+1+7t+7=0$$ $$(t+1)(t^2-t+1)+7(t+1)=0$$

Now it is easy to finish.


Yes it exists. It's called Cardano's Method of Solving Cubics of the form of cubic you posted. This form is called a depressed cubic because it has the $x^2$ term missing.

You can read in detail about it here

But it is not required in your case. You can run the rational roots test and find out that one of the roots can be easily found.


Yes,there is a formula called Cardano's formula but I recommend you not to use it as it is too long to work out by hand and hard to remember too.

As for your equation,

$$27x^3+21x+8=0$$ $$27x^3+1+21x+7=0$$ Factoring,we get $$(3x+1)(9x^2+6x+1)+7(3x+1)=0$$ $$(3x+1)(9x^2+6x+8)=0$$

We get answer as $$x=-\frac13$$

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    $\begingroup$ Interesting, you have a same idea $\endgroup$ – Aqua Feb 16 at 15:10
  • $\begingroup$ I personally think your method was better $\endgroup$ – Jordan Lawson Feb 16 at 15:27

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