Cubic Formula for Depressed Cubic

[@NadiehFan]

(Continued from previous message) Somewhat later, towards the end of the 16th century, Vieta came up with a method to solve cubics with only real roots trigonometrically. Vieta had been working on identities for the sine and cosine of integer multiples of angles and since, using modern notation of course, we have (17) cos 3φ = 4·cos³φ − 3·cos φ this implies, for example, that 4·cos³ 20° − 3·cos 20° = cos 60° = ½. We can conclude from this that x = cos 20° is an _exact_ root of the cubic equation 4x³ − 3x = ½. Vieta hit upon the idea to use this to convert depressed cubic equations into an equation of the form 4x³ − 3x = c by applying a suitable _scaling_ (substitution) of the variable. To see how this works, again, in a modernized version, let's go back to the depressed cubic (1) where p and q are real numbers. As we have seen the solution formulas generate cube roots of complex numbers if and only if (½q)² + (⅓p)³ < 0 and this is precisely the condition for the equation to have three distinct real roots. So, how can we use the triple angle identity (17) to solve our reduced cubic equation? The idea is to first rewrite our cubic equation as (18) x³ + px = −q and then substitute (19) x = r∙cos φ This will give us (20) r³∙cos³φ + p∙r∙cos φ = −q Now we want to get 4∙cos³φ rather than r³∙cos³φ at the left hand side so we divide both members by r³ and multiply both members by 4 to get (21) 4∙cos³φ + (4p/r²)∙cos φ = −4q/r³ If we compare this with 4∙cos³φ − 3∙cos φ we can see that all we need to do now to get the desired expression 4·cos³φ − 3·cos φ at the left hand side is to choose r in such a way that 4p/r² = −3 and therefore we can choose (22) r = 2·√(−⅓p) Note that this value of r is real and positive, since (½q)² + (⅓p)³ < 0 implies p < 0. With this choice of r (21) becomes (23) 4∙cos³φ − 3∙cos φ = −½q/√((−⅓p)³) and because of (17) this gives (24) cos 3φ = −½q/√((−⅓p)³) Now, there is a single value of 3φ on the interval [0, π] which satisfies this, which is 3φ = arccos(−½q/√((−⅓p)³)) but since the cosine is a periodic function with a period 2π we have 3φ = arccos(−½q/√((−⅓p)³)) + 2·k·π, k ∈ ℤ and therefore φ = ⅓·arccos(−½q/√((−⅓p)³)) + ²⁄₃·k·π, k ∈ ℤ. Of course, since the cosine is a periodic function with a period 2π this gives only three different values for cos φ by selecting any three consecutive integers for k. And since we substituted x = r·cos φ and since r = 2·√(−⅓p) and φ = ⅓·arccos(−½q/√((−⅓p)³)) + ²⁄₃·k·π we find using k = 0, k = −1, k = 1 that the equation (1) has the solutions x₁ = 2·√(−⅓p)·cos(⅓·arccos(−½q/√((−⅓p)³))) x₂ = 2·√(−⅓p)·cos(⅓·arccos(−½q/√((−⅓p)³)) − ⅔·π) x₃ = 2·√(−⅓p)·cos(⅓·arccos(−½q/√((−⅓p)³)) + ⅔·π) Note that the condition (½q)² + (⅓p)³ < 0 implies that the absolute value of −½q/√((−⅓p)³) is smaller than 1, which means that the trigonometric solution of the reduced cubic equation x³ + px + q = 0 is always possible if this equation has three distinct real roots.

[@rob876]

This is the best explanation for the cubic formula I have ever seen.

[@sebastiencelma234]

I totally agree with you

[@NadiehFan]

To derive a formula for the solutions of a depressed cubic equation x³ + px + q = 0 there is no good reason now to work with a substitution x = a − b. Historically, this type of substitution, that is, representing the unknown as a difference of two other variables, was used to solve depressed cubics of the type x³ + px = q where p and q are positive numbers. This was done using the identity (a − b)³ + 3ab(a − b) = a³ − b³. Similarly, depressed cubics of the type x³ = px + q where p and q are positive numbers were solved using the identity (a + b)³ = 3ab(a + b) + (a³ + b³) and then the unknown was represented as a sum of two other variables. The reason was that this allowed a geometric interpretation of the equation to be solved, the identities used, as well as the solution method. In fact this is how Cardano proceded in his 1545 book _Ars magna_ in which a solution method for cubic equations was first published. Geometric interpretations of algebraic operations were much more common then, mainly because symbolic algebra as we use it today was only developed about a century later (Vieta, Harriot, Descartes). Cardano says in his book that the method he presents is that of Scipione del Ferro. Since Cardano had had the opportunity to examine a private notebook (now lost) of the late Del Ferro where he gave his solution, it is generally inferred that Del Ferro had arrived at his solution by geometric considerations like those given by Cardano. This geometric approach (completing the cube) is similar to the well known methods of interpreting and solving quadratic equations geometrically by completing the square, which was already used by the ancient Greeks and much earlier also by the ancient Babylonians. Now, let us look at a modern algebraic approach to solving a depressed cubic equation. Generally, if we have a depressed cubic equation (1) x³ + px + q = 0 where p and q are assumed to be real, and if we compare this with the identity (2) (a + b)³ − 3ab(a + b) − (a³ + b³) = 0 then we can easily see that x = a + b will be a root of the cubic equation if a and b _simultaneously_ satisfy (3) ab = −⅓p (4) a³ + b³ = −q Since (3) implies (5) a³b³ = −(⅓p)³ we have both the sum a³ + b³ = −q and the product a³b³ = −(⅓p)³ of a³ and b³ which means that a³ and b³ are the roots of the quadratic equation (6) t² + qt − (⅓p)³ = 0 which are t₁,₂ = −½q ± √((½q)² + (⅓p)³). So, we have (7) a³ = −½q + √D (8) b³ = −½q − √D or vice versa, where (9) D = (½q)² + (⅓p)³ is equal to ¼(t₁ − t₂)² where t₁ and t₂ are the zeros of the quadratic polynomial t² + qt − (⅓p)³ at the left hand side of (6). This means that D is actually the discriminant of this quadratic except for a scaling factor ¼. Since the sign of D determines the nature of the roots of the depressed cubic (1) we can, for practical purposes, also consider D as the discriminant of (1). Now, we have three possibilities. First, if D > 0, then (6) has two real solutions so a³ = −½q + √D and b³ = −½q − √D are real numbers. Let (10) a₀ = ³√(−½q + √D) (11) b₀ = ³√(−½q − √D) where the cube roots represent the real cube roots of the real numbers that are the solution of (6), then evidently a = a₀ is the sole real solution of (7) and b = b₀ is the sole real solution of (8). But these are not the only solutions of (7) and (8). If we designate the two _complex cube roots of unity_ as ω₁ and ω₂, that is (12) ω₁ = −½ + i·½√3 (13) ω₂ = −½ − i·½√3 then a = ω₁a₀ and a = ω₂a₀ are two additional complex solutions of (7). Similarly, b = ω₁b₀ and b = ω₂b₀ are two additional complex solutions of (7). So, there are three possible values for a and three possible values for b. But this does _not_ mean that the _system_ of equations (3) and (4) has 3·3 = 9 solution pairs (a, b) because in accordance with (3) the product ab must equal −⅓p. Evidently (a, b) = (a₀, b₀) is one solution pair of the system of equations (3) and (4) since a₀³ + b₀³ = −q and a₀b₀ = −⅓p. And since ω₁³ = ω₂³ = ω₁ω₂ = 1 it is easy to see that (a, b) = (ω₁a₀, ω₂b₀) and (a, b) = (ω₂a₀, ω₁b₀) are also solutions of this system. Furthermore, Since ω₁² = ω₂, ω₂² = ω₁, it is also easy to verify that non of the other six combinations of one of the three possible values of a and one of the three possible values of b is a valid solution pair (a, b) of the system of equations (3) and (4). So, we have three solution pairs (a, b) = (a₀, b₀), (a, b) = (ω₁a₀, ω₂b₀), (a, b) = (ω₂a₀, ω₁b₀) and since x = a + b is a root of (1) if a and b satisfy (3) and (4) it follows that we have three solutions of our depressed cubic equation (1) which are (14) x₁ = a₀ + b₀ (15) x₂ = ω₁a₀ + ω₂b₀ = −½(a₀ + b₀) + i·½√3·(a₀ − b₀) (16) x₃ = ω₂a₀ + ω₁b₀ = −½(a₀ + b₀) − i·½√3·(a₀ − b₀) Clearly, since a₀ and b₀ are real, this means that (1) has _one real and two conjugate complex roots_ if D = (½q)² + (⅓p)³ > 0. Secondly, if D = 0 but p and q are not both zero, then the quadratic equation (6) has two coinciding (equal) real roots, and consequently, we then have a₀ = b₀ = ³√(−½q). From (15) and (16) it then follows that x₂ = x₃ = −½(a₀ + b₀). So, if D = 0 and p and q are not both zero then (1) has one simple real root as well as one real root of multiplicity two. Since the sum of the roots of (1) is zero, all three roots can only coincide if all three roots are zero, which is true if and only if p = q = 0, reducing (1) to the trivial cubic equation x³ = 0. Thirdly, if D < 0, then the quadratic equation (6) has two conjugate complex solutions t₁,₂ = −½q ± i√(−D), and, consequently, a³ and b³ are then conjugate complex numbers in accordance with (7) and (8). This implies that there are then no real solution pairs (a, b) that satisfy the system of equations (3) and (4). However, if we let a₀ and b₀ be the _principal_ cube roots of the conjugate complex numbers −½q + i√(−D) and −½q − i√(−D) in accordance with (10) and (11) then a₀ and b₀ are themselves complex conjugates as well and (a, b) = (a₀, b₀), (a, b) = (ω₁a₀, ω₂b₀), (a, b) = (ω₂a₀, ω₁b₀) are then again the solutions of the system of equations (3) and (4). Since the sum a₀ + b₀ is real and the difference a₀ − b₀ purely imaginary, it follows from (14), (15), (16) that (1) has _three different real roots_ if D = (½q)² + (⅓p)³ < 0. So, we have the peculiar fact that if the depressed cubic (1) has three distinct real solutions, then these solutions are expressed algebraically by means of cube roots of complex numbers. In fact, if p and q are rational and (1) has three different real roots but none of these roots is rational, then it is _impossible_ to express the real roots of (1) algebraically without using cube roots of complex numbers. Any attempt to extract the cube roots of those complex numbers will only result in a cubic equation which is equivalent with the cubic equation we were trying to solve in the first place. This famous catch-22 which had Italian algebraists in the era of Cardano running around in circles (no trigonometric pun intended) is known as the _casus irreducibilis_ or irreducible case. (Continued in next message)

[@bruhifysbackup]

Amazing!

[@_emi_E]

I devised a procedure for working the 3 roots of a (depressed) cubic equation using (i) any suitable one of the trigonometric identities or its hyperbolic version for cos3x, sin3x, cosh3x or sinh3x, & (ii) the complex identities of their inverses - expressed in log (derived from their exponential form) The procedure always work, whether or not roots are all real or a complex conjugate pair exist.

[@dougaugustine4075]

I continue to be amazed almost every day.

[@domedebali632]

Same here. Sure you have an interest in math

[@TheLukeLsd]

Eu já conhecia mas foi a forma mais simples e intuitiva de deduzir a fórmula que eu já vi. Parabéns

[@jamestopfaff]

Amazing video bud, I love your content. Hopefully this cubic overcomes its depression soon

[@KushalRoyal-r3t]

Very nice sir!! Once I was trying to find roots of a cubic equation and I depressed(randomly) it at that time I dont know what is a depressed cubic equation, and i even dont know how to solve that. Once I saw this video on my recommendation I was very happy by seeing its thumbnail and reading the title. Then i watched the whole video and now i know how to find roots of cubic equation. I am very happy and Thankful sir❤!!

[@vadimtokman123]

Wow, I could not believe, that you explained that simple, that even I understand...... Thank you!!!!

[@anaycoding6594]

Very clearly explained 👍

[@holyshit922]

Maybe it is a good idea to generalize it for quartic as homework (Generalization can be found in Euler's book Vollständige Anleitung zur Algebra)

[@kragiharp]

Wow, how interesting! I have never heard of this before and never thought about it. Thank you so much! ❤️🙏

[@mikezulu1349]

Beautiful! You are an excellent teacher. 👏

[@mathtronomy]

Cool, you teach really simple and easy to learn. thank you

[@marilynman]

"That was too close, but we survived" Every time I try to squeeze everything in the confines of a sheet.

[@PsYcHoCI2usHeI2]

Man, I wish this had come out months ago. There are hardly any videos that clearly explain the Cardanos formula.

[@ThePhotonMan110]

Could you do a worked example of this for p and q being some integer values? I'm confused how to use the final equation to find all of the roots instead of just one

[@williamperez-hernandez3968]

Historical Fact: Tartaglia is not his real name, he was Nicolo Fontana. Tartaglia was a nickname given to him because of his speech impairment, stuttering. In Spanish, people with this disability are called Tartamudo. Suppose in Italian it is similar, thus the "Tarta" at the beginning of the word. Scipione del Ferro is mentioned as the first to achieve solving the depressed cubic, but didn't publish it. Apparently he showed it to his student Antonio María Fior, who debated with Tartaglia as whom had the distinction of being the first.

[@bowlineobama]

You are very good teacher. Thank you. But, I don't understand why you are letting X= a - b. What is the reason ?

[@justekiara1953]

Merci. We must revise it several times to master it. Thanks.

[@lukaskamin755]

Wow this way of achieving the Cardano-Tartaglia formula. Really makes sense. I just wanted you to explain how we can derive 3 roots out of one formula. That's the most shady part. I've several various ways to achieve three other roots, without explanation wherw they come from. To cut it short there needs to be a separate video on applying this formula to get all 3 real solutions when they exist, and how to know how many solutions will be. In most of if the cases the solutions ended in approximate calculation of thise roots on calculator and that looked kinda disappointing. I mean if the equation is sure to have 3 integer roots( you may even construct it for purpose, knowing the foots from the beginning), to achieve that integer roots from Cardano formula

[@flightyavian]

There is a method for extracting all three roots so in the standard formula, you have two cube roots and then -b/3a at the end anyways, what you want to do is find the simplest form of what's inside both cube roots and then, convert the numbers inside both roots to polar form you'll either have a complex number or a real number, with real numbers the angle is just zero in the complex plane so once you find the modulus of what's inside both cube roots, you take the cube root of the modulus in each root, and multiply it by cis(θ), where θ is equal to (φ+2kπ)/3, where k is equal to either 0, 1 or 2 depending on which root you want if you pick 0 in one cubic root you pick 2 in the other, and vice versa and if 1, you pick 1 for both. this gives you three different combinations for the three different roots Edit: phi is the angle in the polar form of the number inside the cube root you will have to find two different moduli and two different angles for both of the cube roots in most cases if it's a complex number, the angle is just arctan(b/a) when the number is in the form of a+bi the explanation for this is that this basically how you find the three values of a cube root in general, and I'm not really sure why when you pick 0 in one cube root you must pick 2 in the other but for 1 they must be the same I'm not sure how you can deduce whether or not there will be 3 real solutions or only 1 without just immediately solving the thing right away finding one solution and then analyzing the quadratic is one way, but somehow I don't think that's what you want I think I read somewhere that if all the solutions are integers you will always have a complex number under the cube root beyond that gl

[@fioscotm]

Great Video!!!!! Finally I understand the method to solve a cubic (even if it only works sometimes). Can't wait for the next video!

[@KPunktFurry]

looks funny! 0:32 Can't we just do +0*X² and then insert it to the left, that wouldn't change the value? 1:04 I am fascinated by your happyness and interest in math and unfortunatly i don´t knowe this formular but i will surely try to memorize it 1:50 ok lets show the formular 5:15 ok we rewrite the equantion and move a cubble of variables arround 6:45 that is why i love the midnight formular or abc-formular 8:16 ok i would simplify the fractions to (27a²+4P³)/(4*27) 13:12 plusminus and minusplus ist the same! 14:02 but can´t you simplify that any more? 14:08 of cauce i will left a comment in the comment section! see you again K.Furry

[@lumina_]

that was really easy to follow thank you!! other videos on this topic scared me so i never watched ahaha. excited to see the next vid on how to depress any cubic

[@aashsyed1277]

i reallly like your chalk! which one do you use???

[@tomctutor]

Apparently all cubic polynomials can be reduced to the "depressed" form shown here; it was delFerro et. Fior whom discovered the general solution in early 16th century (for the type in the example). These early methods struggled understanding square root of negative numbers and it was later that Bombelli factored in these imaginary roots.

[@uwuowo7775]

That intro is a banger

[@skwbusaidi]

I usualy let x=a+b And solve quadratic that it roots are a^3 AND b^3 and take the cube root and sum then and it does not mater which one is a and which one is b because we want the sum a+b This is one root then do division to get quadatic in x and get the other two roots

[@kassuskassus6263]

Sir, you are the Rambo of the mathematics ! 😀😀😀

[@misterj.a91]

I think it's easier to do the proof each time than to memorize it.

[@Yamcha1717]

a mnemonic technique is to remember that in second place comes (q/2)^2 and in third place comes (p/3)^3

[@cadenpink316]

This is my preferred strategy. I think human brains are so diverse, that everyone should find what works for them.

[@flightyavian]

It's not so bad if you break up the memorization into chunks I like to personally remember ³√(p+√(p²+q³)) + the same but change the middle sign, -b/3a Where p is equal to -b³/27a³ +bc/6a² - d/2a and q is equal to -b²/9a² + c/3a sure, still a lot but you can easily mash this into a calculator by just finding the values of p and q Edit: I mean it doesn't really matter, if I encounter a cubic I'll just solve it using my calculator, no using formula but yeah

[@darcash1738]

Set some variables for yourself to make it easier. K = -q/2 M = q^2/4 + p^3/27. Note how once you start this term which is squared is when it also adopts another sqrt. The denominators are nice because 4 = 2^2 and 27 = 3^3, so these powers actually match the numerator. x = cbrt[K +/- sqrt(M)] + cbrt[K -/+ sqrt(M)] Start with this simplified form and then write K and M. Whatever comparisons you make between elements within the formula will help. It is important to remember the signs, exponents, and coefficients. You can break these apart or combine them together into a unit of your memory depending on what makes the most sense to you I personally think of it as q q p, ascending in power. I break down to K and M to make the cube root and sqrt more distinct

[@jasimmathsandphysics]

Just do it so many times that you remember each step

[@holyshit922]

To complete video of course depressing cubic is missing but it would be good idea to show how to get all solutions using cube root of unity It would be good idea to consider casus irreducibilis separately and derive trigonometric solution to this case Tartaglia - i wouldn't read g in this name

[@NadiehFan]

You may want to check out my comments on this video. I've already discussed what Prime Newtons should have discussed in this video.

[@holyshit922]

@@NadiehFan Yes I agree that this what is in your comments should appear in video

[@holyshit922]

@@NadiehFan If you record video about cubics it will be better than Newton's video

[@PrimeNewtons]

Thanks

[@holyshit922]

@@PrimeNewtons Look at NadiehFan comments Two of them are really helpful for this topic

[@emad3241]

now i'm convinced that i'm a depressed cube

[@felipelleite7285]

What amazing video. I wish I have found your channel ealier. Keep the great work =)

[@mentooo5709]

This was good fun, great explanation

[@ruud9767]

Wonderful. Thanks!

[@arthurtoso2582]

incredible

[@sovietwizard1620]

I looked into this earlier, but then I realized it was far more unnecessary than to just use a decimal root approximation. It would take me like 20-30 minutes to solve 1 cubic and the answers I would get are ridiculus. And for the 3 roots where we had to use polar coordinates and then undo them, I got even more confused and didn't even manage to solve 1 problem for that. Luckily 2 roots actually have a small other formula which you can use on wikipedia (but 2 roots are rare anyways)

[@lawrencejelsma8118]

I think your x1 and x2 are the complex conjugate pair roots. The real root probably doesn't exist to make an x^2 term not appear (and why you only found two roots to the depressed cubic equation). I was wondering how an x3 real root wasn't shown in this proof but that comes from another KZbin creator solving ax^3 + bx^2 + CX + d; where, a = 1 and b = 0 reduces to yours and he shows the complex number root pair formula without what the real root additional solution would be in his also. I would prefer finding polynomial roots by a Muller's Method algorithm to either find the real root dividing out leaving a solvable quadratic polynomial or the quadratic polynomial leaving only the real root should it exist factor. The reason why I prefer iterative mathematics in polynomials n >= 3 is for electronic circuit design reasons of only capacitors, resistors and inductors being the only real components for up to second order transfer polynomial function reasons. A transistor operating in its active region models as a resistor, capacitor and transistor with additional current source modelling (extra input excitation as the voltages) and no other real components have been found to expand to third, fourth, etc. transfer function polynomials without doing analog feedback of output signals to input signals in circuit designs. The output to input feedback of signals without any other physical components to come up with expand the transfer function polynomial beyond a second order of T(x) order n>= 3 in powers of x as time considerations.

[@arma-geddon-13]

you're very cool!

[@VineetaSingh-m5j]

Sir I am from India and I was taught that quadratic formula was invented by an Indian mathematician named sridhara and it is know as sridharacharya formula Sir what,s your opinion on this

[@CalculusIsFun1]

It’s unfair to say only one person “invented” the formula. Many different cultures independently discovered it over thousands of years. Greeks, Babylonians, Persians, Indians, Europeans, etc. No one person or group discovered the formula.

[@KrishnaKantaBehera-rg9gk]

Yo bro Even if you say this.... I am saying for your sake In literal meaning, it is Sridharacharya's formula not Sridharacharya formula meaning it is his formula not the formula is named after him And also the fact that it is a formula for cubic expression so it is called cubic formula If I expand the name it would be "Sridharacharya's cubic formula" In another civilization this formula may be assigned to another great mathematicians we may not know about..... But my point is.... This is a cubic formula for cubic expression and it has been founded by many people So you can never be sure who founded it

[@pojuantsalo3475]

Nice one! 👍

[@Arkapravo]

Good one!

[@machoodin5172]

But the cubic function can have up to three real roots and the formula can have only one How do I find the other ones???

[@NadiehFan]

See my comments on this video where this is explained.

[@machoodin5172]

@@NadiehFan Thanks!

[@AlmondAxis987]

The problem is, there are actually 2 solutions by this formula

[@jayy7842]

Yay!

[@boguslawszostak1784]

I prefer x=a+b

[@ThoseInterestingStories]

How would that even work

[@boguslawszostak1784]

@@ThoseInterestingStories works the same exactly. (a+b)^3=(a^3+b^3)+3ab(a+b) x^3= -q -px (a^3+b^3)= -p 3ab= -p ....

[@sinox5]

This formula can output two roots, but cubics can have three. Where's the third one?

[@tomctutor]

Multiply the solutions by primitive cube roots of unity (-1 ∓ i/√ (3))/2

[@Th3OneWhoWaits]

Pretty sure it's domain based. Real numbers give us two, complex gives us more.

[@NadiehFan]

He did not find the expression for two roots, just the expression for _one_ of the roots. See my main comments on this video for a full explanation.

[@alexanderwelch7210]

You have to multiply an and b by alternating cube roots of unity. So the solutions are a-b, aω-bω², and aω² -bω (remember he said the formula yields the same result regardless of which one is positive and which is negative)

[@bhagyashrigadekar8618]

Solve something examples based upon it try mostly irrational ones 🗿✌️😎

[@holyshit922]

kzbin.info/www/bejne/onnIZ5yfmtateNk Just for reminder We should come back to this system p = 3ab , -q = a^3 - b^3 and use cube roots of unity to get all solution to this system and then to cubic equation Casus irreducibilis is also worth to consider because in this case complex cubic radicals lead us to real solutions With de Moivre's theorem we can get trigonometric solutions from this case

[@MataniMath]

This formula does NOT apply to the equation x³ - 7x + 6 = 0; p = -7 & q = 6. Let n = q²/4 + p³/27 ----> n = 9 - 343/27 < 0 and the sqrt(n) is a complex number. But the solutions of x³ - 7x + 6 = 0 are x = -3, x = 1 & x = 2. All of them are real number. It means this formula is not valid or invalid

[@NadiehFan]

No, you are wrong. The formulas to express the roots of a depressed cubic x³ + px + q = 0 are always valid. However, if (¹⁄₂q)² + (¹⁄₃p)³ < 0 then the cubic has three distinct real roots, _but_ these can then only be expressed in terms of the coefficients p and q using cube roots of complex numbers. For p = −7, q = 6 we have D = (¹⁄₂q)² + (¹⁄₃p)³ = 3² + (−⁷⁄₃)³ = 9 − ³⁴³⁄₂₇ = ²⁴³⁄₂₇ − ³⁴³⁄₂₇ = −¹⁰⁰⁄₂₇. Therefore, √D = i·√(¹⁰⁰⁄₂₇) = i·10/(3√3) = i·¹⁰⁄₉√3. So, one of the roots of your equation is x₁ = ³√(−3 + i·¹⁰⁄₉√3) + ³√(−3 − i·¹⁰⁄₉√3) for which we can write x₁ = ¹⁄₃·(³√(−81 + i·30√3) + ³√(−81 − i·30√3)) Now, you can verify by cubing that we have ³√(−81 + i·30√3) = 3 + i·2√3 ³√(−81 − i·30√3) = 3 − i·2√3 so we get x₁ = ¹⁄₃·(3 + i·2√3 + 3 − i·2√3) = ¹⁄₃·6 = 2 which is indeed one of the roots of your cubic equation. The other two roots x₂ and x₃ are obtained by multiplying the cube roots in the expression for x₁ by ω = −¹⁄₂ + i·¹⁄₂√3 and ω² = −¹⁄₂ − i·¹⁄₂√3 _in either order_ so we get x₂ = ¹⁄₃·((−¹⁄₂ + i·¹⁄₂√3)(3 + i·2√3) + (−¹⁄₂ − i·¹⁄₂√3)(3 − i·2√3)) = ¹⁄₃·((−⁹⁄₂ + i·¹⁄₂√3) + (−⁹⁄₂ − i·¹⁄₂√3)) = ¹⁄₃·(−9) = −3 and x₃ = ¹⁄₃·((−¹⁄₂ − i·¹⁄₂√3)(3 + i·2√3) + (−¹⁄₂ + i·¹⁄₂√3)(3 − i·2√3)) = ¹⁄₃·((³⁄₂ − i·⁵⁄₂√3) + (³⁄₂ + i·⁵⁄₂√3)) = ¹⁄₃·3 = 1 As you can see, the formulas give the correct roots of your equation.

[@MataniMath]

@@NadiehFan You're right. I'm jumping to conclusions. Thank you very much

[@knaeve]

aka Cardano method.

[@RikiFaridoke]

You are wrong sir, it is should have tree root of qubic equation, there is one real root, and two complex roots of qubic equation, so the all of roots is conquered to Descartes sign basic laws of polynomyal.

[@tomctutor]

Generally it will result in only one real solution, e.g. x^3-6x+9 =0 will result in x= -3, the other two here are complex got by further multiplying -3 by primitive complex cube root of unity = (-1 ∓ i/√ (3))/2 . Or try using long division using the real root; so x^3-6x+9 = (x+3)(x^2-3x+3) where the quadratic factor is irreducible, hence has complex roots. (It may also result in complex root if p is large enough negative).

[@RikiFaridoke]

@@tomctutor yes sir, oftenly, if we using this method, we Will usually get two complex roots in equation.

[@cremath]

depressed?? 😂