[Algebra] Is there any cubic formula?

In yesterday's video, we tried to solve an inequality of rational functions (functions in fraction form for both numerator the top part and denominator the bottom part being polynomials) but ended up with one of the factors a cubic polynomial that was not factorable with integer coefficients. It led us to the following question: is there any way that we can solve such a cubic polynomial?

In math class, we learn that for any quadratic equation, there exists a general solution called the quadratic formula; today, we are trying to figure out a general solution for an arbitrary cubic equation, a.k.a. the cubic formula, by using Cardano's method.

The remaining derivation of the cubic formula is in the next video:

[klepikovmd]

Awesome video, one of the best explanation of the cubic formula.
Can you please do a quartic?

[ninoneillopez]

Since you already have that cubic formula, can you give us some examples of cubic functions using your derived formula? For instance, 4x^3 - 108x^2 + 720x - 1400. Can you solve this using your formula? Thanks in advance :)

[davidbrisbane]

I played a bit more with the cubic equation and found that in some cases it is easy to algebraically transform a real cube root derived from a cubic equation formula into something much easier to work with.

The real root of the equation x³ + 3x - 4 = 0 according to the cubic equation formula is ∛(2 + √5) + ∛(2 - √5) and it is relatively easy to show algebraically (without reference to the original cubic equation) that ∛(2 + √5) + ∛(2 - √5) = 1.

[keescanalfp]

Very nice algebra!
Now during your rolling out of it i began to wonder what would happen with the formulas and the derivation - not with the solutions, of course - when we would start instead of a, b, c, d, with a, 3b, 3c, d, so from
ax³ + 3bx² + 3cx + d = 0
and likewise in the middle with
X³ + 3pX + q.
?

[WahranRai]

5:31 I dont like your substitution : the correct one is : X = x+ t ---> develop in the equation, group the terms and set the coefficient of x^2 = 0 ---> t = b/3a

[nasrullahhusnan]

Viewer may wonder how you come up with transformation of substracting x with b/3a. It will confinience to follow the explanation if we begin with
f(x)=ax³+bx²+cx+d
The second derivative of f(x) is 6ax+2b. Equating it to zero gives x=-(b/3a), the absica of the inflection point of f(x). Let g(x)=f[x-(b/3a)]. g(x) is f(x) being moved in such a way that the inflection point is in the y-axis. Upon simplification we get g(x)=a(x³+px+q), where p and q are as you have shown. Crossing point of g(x) with the x-axis is g(x)=0 --> x³+px+q=0. Thus the crossing point is the roots of x³+px+q=0. The roots of the equation may be one, two, or three, depending on p and q.

[calvinjackson]

Surely you know the cubic formula was discovered in the 1500s. It's been around for quite a while.

[Lepl_Acie]

Please can you make a video for general formula of 3rd and 4th degree ( in equation ) i'm confused because we have many methods !!!

[ingiford]

That gets one root. once you factor out that root, you have a quadratic left and you can get the other two.

[hocinemerah]

How solving the quintic equation (5 drgre)

[davidbrisbane]

I tried an example to find a real cube root.

I picked a³ + a² - 36 = 0. It is a cubic, so it has at least one real cube root.

Let t A = (a + 1/3) to find the depressed cubic,
which is A³ - A/3 - 970/27 = 0.

Now let A = u + v with A³ - A/3 - 970/27 = 0, and let u, v ∈ ℝ.

⇒ (u + v)³ - (u + v)/3 - 970/27 = 0
⇒ (u³ + v³ - 970/27) + (u + v)(3uv - 1/3) = 0

So, setting u³ + v³ = 970/27 and uv = 1/9 we can solve the depressed cube and hence the undepressed cubic.

Let U = u³ and V = v³ and we obtain
U + V = 970/27 and UV = 1/9³ = 1/729

Now if we consider (Z - U)(Z - V) = 0
⇒ Z² - (U + V)Z+ UV = 0
⇒ Z² - 970/27Z+ 1/729 = 0
⇒ Z = [970 ± √(940, 896)]/54

⇒ U = ((970 - √(940, 896))/54 and
V = ((970 + √(940, 896))/54

⇒ u = ∛(970 - √(940, 896))/54) and
v = ∛(970 + √(940, 896))/54)

⇒ A = ∛(970 - √(940, 896))/54) +
∛(970 + √(940, 896))/54)
⇒ A = (6 decimal places)

But, actually, A = 10/3, because we know
a³ + a² - 36 = 0 has a real root when a = 3
⇒ A = (a + 1/3) = (3 + 1/3) = 10/3.