# Roots of the Cubic equation

### The Cubic formula:

$$ax^3+bx^2+cx+d=0$$

With under the following conditions:

1. $$a \neq 0$$
2. $$a,b,c,d \in \Bbb{R}$$

We can derive the following formula as a the root of $$x$$:

$$u= \sqrt[3] {{bc\over 6a^2}-{d\over 2a}-{b^3\over 27a^3}+\sqrt{({bc\over 6a^2}-{d\over 2a}-{b^3\over 27a^3})^2+({c\over 3a}-{b^2\over 9a^2})^3}}$$

$$v= \sqrt[3] {{bc\over 6a^2}-{d\over 2a}-{b^3\over 27a^3}-\sqrt{({bc\over 6a^2}-{d\over 2a}-{b^3\over 27a^3})^2+({c\over 3a}-{b^2\over 9a^2})^3}}$$

$$x_1= u + v-{b\over 3a}$$

But the proof of derivation of the formula mentioned above was only limited to 1 root. In addition, that formula had no complex conjugates. However as we know, there must be 2 other roots which includes complex conjugates in their formulas.

In reference, Wikipedia: Cubic equation also says that there should be 2 other roots at maximum.

So, altogether the 3 roots are:

$$i= \sqrt {-1}$$

$$\omega = -{1\over 2} + {{\sqrt 3}i\over 2}$$

• $$x_1= u + v-{b\over 3a}$$

• $$x_2= {\omega }u + {\omega}^2v -{b\over 3a}$$

• $$x_3= {\omega }^2 u + {\omega}v -{b\over 3a}$$

And unfortunately, I didn't find or know know the proof of any other 2 roots, i.e $$x_2$$ & $$x_3$$.

So would you please show me the proof of the other 2 roots of Cubic formula?

Note: Please, no synthetic division. I need proof by formula

• Every general cubic can transform to reduced cubic $y^3-3py-2q=0$ by substituting $x=y-\dfrac{b}{3a}$. If once accept the identity $$a^{3}+b^{3}+c^{3}-3abc \equiv (a+b+c)(a+\omega b+\omega^{2} c)(a+\omega^{2} b+\omega c)$$ and replacing $a, b, c$ by $y, -u, -v$ respectively, $y^{3}-3uvy-(u^{3}+v^{3})=0 \implies y_{k}=u\, \omega^{k}+ v\, \omega^{2k}$ for $k=0,1,2$. Note that $\omega^4=\omega$ and see more in another answer. – Ng Chung Tak Aug 12 at 3:09
• Further tricks, try to simplify $u^3+v^3$ and $uv$ from your expressions first. – Ng Chung Tak Aug 12 at 3:20
• Wow, @NgChungTak That helps & I hope that should be illustrating the answer of this question. Thanks a lot – Prince Khan Aug 12 at 3:40

Use polynomial long division to divide $$ax^3+bx^2+cx+d$$ by $$x-x_1$$. This will give you a quadratic, and from there you can find the two remaining roots.

You can do this systematically, using the usual steps for division. If the polynomial is relatively simple, then an alternative would be to write out $$ax^3+bx^2+cx+d = (x-x_1)(ax^2+px+q),$$ then start matching coefficients. For example, looking at $$x^2$$, we have $$b = p-ax_1$$, and so on.

• That's a nice idea, but however, I said proof by formula. Although I didn't mentioned the synthetic division, I'll edit my question so that no more confusion like this arose. – Prince Khan Aug 12 at 2:44
• @PrinceKhan You can easily derive that formula by carrying out division on the general form. – Théophile Aug 12 at 14:12
• Such as? I said I need proof BTW, everyone can do that arithmetically – Prince Khan Aug 12 at 15:37

Starting with the "depressed" cubic:

$$y^3 + py + q = 0$$

If you don't have a diminished cubic you can substitute $$x = y - \frac {b}{3a}$$ which will eliminate the $$bx^2$$ term from the original cubic.

Next we do a similar substitution:

$$y = z-\frac {p}{3z}$$

$$(z - \frac {p}{3z})^3 + p(z-\frac {p}{3z}) + q = 0\\ z^3 - pz + \frac {p^2}{3z} - (\frac {p}{3z})^3 + pz-\frac {p^2}{3z} + q = 0\\ z^3 - (\frac {p}{3z})^3 + q = 0\\ z^6 + qz^3 - (\frac {p}{3})^3 = 0\\ z^3 = -\frac {q}{2} \pm \sqrt {(\frac {q}{2})^2 + (\frac {p}{3})^3}$$

We can choose $$z^3 = -\frac {q}{2} + \sqrt {(\frac {q}{2})^2 + (\frac {p}{3})^3}$$
with $$\frac {1}{z^3} = \frac {-\frac {q}{2} - \sqrt {(\frac {q}{2})^2 + (\frac {p}{3})^3}}{-(\frac {p}{3})^3}$$
Or, $$-\frac {q}{2} - \sqrt {(\frac {q}{2})^2 + (\frac {p}{3})^3} = (-\frac {p}{3z})^3$$

Note that $$\omega^3 = 1$$ has $$3$$ roots
$$\omega_1 = -\frac 12 + \frac {\sqrt 3}{2} i\\ \omega_2 = -\frac 12 - \frac {\sqrt 3}{2} i = \omega_1^2 = \frac {1}{\omega_1}\\ \omega_3 = 1$$

$$z = \omega {\sqrt[3]{-\frac {q}{2} + \sqrt {(\frac {q}{2})^2 + (\frac {p}{3})^3}}}$$

with $$z$$ equal to any of the values of $$\omega$$ above, and

$$y = z - \frac {p}{3z} = \omega {\sqrt[3]{-\frac {q}{2} + \sqrt {(\frac {q}{2})^2 + (\frac {p}{3})^3}}} + \frac {1}{\omega}{\sqrt[3]{-\frac {q}{2} - \sqrt {(\frac {q}{2})^2 + (\frac {p}{3})^3}}}$$

If $$(\frac {q}{2})^2 + (\frac {p}{3})^3 >0$$ there is only one real root.
If $$(\frac {q}{2})^2 + (\frac {p}{3})^3 <0$$ the complex terms ultimately cancel out and there are 3 real roots.

• How did you changed that? $\frac {1}{z^3} = \frac {-\frac {q}{2} - \sqrt {(\frac {q}{2})^2 + (\frac {p}{3})^3}}{-(\frac {p}{3})^3}$ – Prince Khan Aug 12 at 10:03
• $\frac {1}{z^3} = \frac {1}{-\frac {q}{2} + \sqrt {(\frac q2)^2 + (\frac{p}{3})^3}}\left(\frac {-\frac {q}{2} - \sqrt {(\frac q2)^2 + (\frac{p}{3})^3}}{-\frac {q}{2} - \sqrt {(\frac q2)^2 + (\frac{p}{3})^3}}\right) = \frac {-\frac {q}{2} - \sqrt {(\frac q2)^2 + (\frac{p}{3})^3}}{-(\frac p3)^3}$ – Doug M Aug 12 at 10:28
• Hey thanks dude, that helped – Prince Khan Aug 12 at 14:09