Finding a cubic formula for roots of cubic equations

Question

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.

Answer 1

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!

Answer 2

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.

Answer 3

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.